hello my dear are you all ready are you all excited my dear get bat boys and G bat girls okay very good so malam Gigi nice to see you arut San mesh Kumar Arjun Rd adya chukes sanit har so nice to see you the chap Sol previous year questions pqs okay so I have covered literally everything hour and half hours try to help you okay so in order okay so chapter physics I will teach you second chapter electrostatic potential and capacitance and chemistry solution so solution I will teach you maths is it's nothing okay okay so I will teach you all the three subjects so de I hope all of you are with me it will give them confidence and it will give them some hope okay so Target 48 sessions toic if you're watching it later and practice nothing will beat hard work hard work sincerity and consistency nobody can stop you from success that's the truth of Life nobody can change that so if you are sincere if you are consistent if you're hardworking nothing can stop you from reaching whatever Target you set for yourself okay okay so that's my job you focus on studying you focus on solving very good yesterday's video is not showing for me I have thought four hours we will start soling you might have seen this earlier and return we are going to work together as a team okay and already subscription smash the like button like button so let's get started recent question recent question from this chapter I want everybody to try it out a uniform electric field is represented as 3 into 10^ 3 Newton per Kum I find the electric flux of this field through a square of side 10 cm when the plane of the square is parallel to y z plane okay I will also try to help you okay so first of all what is electric flux electric flux because of this field so electric flux or formula p is equal to e do a right so so I'm going to solve a part A Part A when the plane of the square is parall to y z plane so my flux is equal to e Vector e Vector is 3 into 10^ 3 I okay dot a vector what is area Vector y z PL vect can anybody let me know please tell me quickly very good mes AR very good answer sir plus xus X I don't know okay whether it is plus X or minus X we don't know but yeah we can take it plus X so I or x-axis direction is correct so it will become along xais is that so y that plane now see x axis y axis is that plane in the plane okay so I will use that book okay so I will use the book I okay okay so this is what is y z plane I hope all of you are clear with it so Square VOR will be perpendicular to that okay area Vector will be perpendicular to that so perpendicular to that y y plane okay I hope all of you are clear so now area is along X AIS we'll take it along x-axis so what is the area area is square square area is nothing but side into side so it is going to be 10 cm into 10 cm pasang always keep it in SI units always keep it in SI units so area is 1 by 100 m Square so the answer is 30 okay so answer is 30 Newton M Square per Kum okay so I hope all of you are clear everybody happy great abish level abish sham AR very good is I hope all of you are clear y yis AIS okay great question sir the normal to the plane normal to the plane are vector correct normal to the plane area Vector it makes an angle of 60° with x-axis so xaxis the electric field Vector so indect I hope all of you are able to understand so indirectly is giving you information normal to the plane is area Vector x-axis is electric field Vector yeah electric field is along the x-axis so indirectly is giving you day angle angle between area and electric field is 60° so now it's very simple no now what's the answer we know 5 is equal to e vector. a vector so e Vector is what 3 into 10 ^ 3 I cap and a vector is what a vector is going to be it makes 60° so e do a is e a e a by 100 by 100 e a COS Theta so COS of 60 Dee so e do a is magnitude of e magnitude of a COS Theta and Dot product is What DOT product is e do a is nothing but e a COS Theta e magnitude is this one and a magnitude is side of a square into side of a square so square area side into side side is 10 cm converted to meter cimeter to cim to met divided 100 so I want to keep it as meter answer so this is equal to 15 newton M Square per Kum so that's the answer I want you all to be fast okay are you all clear everybody happy very good very good super I'm so glad you guys are all able to understand exam pad exam so XIs yis xyy X it is having XIs and yis XY XIs yis in the plane soe it's a 2d it's a two dimensional figure okay plane is a two dimensional figure okay are you all clear very good outside axis are all perpendicular to each other x axis is per perpendicular to Y axis Y axis is perpendicular to Z axis Z axis is perpendicular to xaxis so all of them is perpendicular to each other so they are mutually perpendicular so is going to come outside okay will be outside are you all clear y it is xais correct normal normal so normal it will be perpendicular to your Square so square square it is par to y z plane so y so it ismal is going to be along X Direction I took along XIs Vector are you all happy can you all think about it everybody it's okay let's invest sometime I will complete the session no matter what no matter what I will complete the session but think about it no X is at plane so X plane will be like this so this is your X plane so x x soe are you all clear are you all happy very good I hope you all enjoyed it now let's move forward let's go to the next question okay come on guys I I want all of you to try it out the magnitude of electric field due to a point charge at a distance of 4 m is 9 Newton per Kum for the same charge for the same charge the electric field of magnitude 16 Newton per Kum will be at what distance okay everybody try to solve it I want you guys to be fast venzo yes yes venzo now okay so are you all ready what is the correct answer come on guys I want you all to think about it very good v v is absolutely right chukes check pukes please check it a is Right siril very good sill abishek Zia Rahul very good nandana Super nandana gopika Dave Amir haran Mohammad very good Mohammad aan super yeah chukes check chukes Mohammad it is three 3 m is the answer check corre okay so due to a point charge electric field electric field to ele to a point Char is KQ by r² I hope all of you are aware so dist yes electri so now we Canute what is KQ KQ KQ is 16 into 9 so KQ is nothing but 16 into 9 sare did you all enjoy it are you all happy with it my dear boys and girls super very good you guys are brilliant ramama very good ramama Pro Gamer Vidya arut San great job guys all of you are absolutely right Pang did you understand this is the way to solve it get don't waste too much time okay let's go forward so everybody try to solve this3 everybody please try to do it quick quick quick I want you all to be fast fantastic as AIA superb siril very good s keep trying guys in the pbd upload week I will upload it in a common place where all of you can download very good so Bangla are you all done shall we solve this question very good you can just use your logic and just tell the answer it is not too difficult okay so solve very good abishek very good super soang you have you have two identical charges so electric field here you have identical charges it is both positive or both negative you cannot have one positive and one negative be careful unnecessarily you lose marks okay so two tical point charges so we have at origin charge and at a comma Z at a comma 0 I have another charge okay are you all clear with this everybody happy okay everybody cool with this I want you guys to be fast come on guys orig and I have minus1 plus 4 charge can the electric field be zero in between them at any point if so find that point in the charge line join the line joining both these charges okay in between the charge at some point you have the field to be zero is it true can it be can it be possible never in the in the case attract positive charge will repel so because of both the charges the field will be in the same direction are you guys able to understand so in the charge in the line joining both of them the field will be to the left you negative charge will attract and positive charge will repel did you understand it so if you have one negative and one positive it will be somewhere here it'll be somewhere here yes negative charge will attract and positive charge will repel to the left side of minus1 charge there is a possibility of your electric field being zero con Concepts I'm teaching you try to understand minus one will attract four right you can have field zero so because four will dominate it's a bigger charge and you're keeping it near how is it possible so it will be only here concept I hope you understood they can ask you like this they'll give you one negative charge uh one positive charge in they will divide into three regions they will divide it into three regions they'll ask you answer is one so in the answer one I hope all you are able to understand so please be ready for it okay this is a very simple question charges both are opposing yes sir that will also field becoming zero is possible the field becoming zero is POS are you all clearid because it's a symmetric figure it's a symmetric system q q same charge same value they are separated by a distance a so where will the field be zero definitely at the middle so answer is answer is Middle answer is a that is when I will be upset okay so now let's try to solve this everybody listen carefully charge it is at a distance of xge it is at a distance of a minus x correct do you agree are you all happy P the field is zero I'm assuming at a point P the net electric field is zero at a point P the net electric field is zero so the point I don't know so charge at distance X charge at distance of distance minus X distance a so a minus X okay so the field at P because of mod charge yo the field at P because of modal charge is equal to KQ by x² assuming this is a positive charge assuming this is a positive charge okay and the field at P because of the r the charge is equal to again KQ by a - x the whole s minus I assuming it's a positive charge super so field N Net field at p is M so net field at p is M net field at p m so the field at P because of first charge the field at P because of second charge x² should be equal K by Aus x² they will get subtracted I minus I so subtract I should get zero so if I subra and it zero that means they are equal if they're equal now you can solve it KQ KQ cancel a - x² is equal to x² a a - xal to x a isal 2X x is equal to a by 2 procedure procedure I hope all of you are able to understand so shall I move forward I hope you are all happy with this so please please okay very good so yeah now let's move forward my dear boys and girls let's proceed to the okay which is here yeah let's see how many of you are able to solve this okay so yeah no no problem you can you can continue from here I will solve all of it session hour and half hours I want everybody to attend it completely till the very end okay super I want everybody to try it out an electron experiences a force in the in a electric field e then the electric field e is how much how much is the electric field so pangla what is the force in an in electric field what is the force experienced by a charge in electric field very good mishwar very good siril GOP trishant Sanai very good Dave Ru super AR right Cham abishek very good abishek super P Amir Zia great very good I willach you the force experienced by a by a charge in an electric field is Q into e moving when you have both electric and magnetic field how will a charge move that's a very nice concept I will teach you that very well probably in the second or third week of December okay so right now this is a force you force experienced by a moving charge by a charge in an electric field not a moving charge Force experienced by a charge in an electric field okay great so force force 1.6 into 10^ minus 16 newon in Direction IAP direction that is equal to charge charge electron electron - 1.6 into 10^-9 please remember that electron is negative charge so minus what is the charge on electron charge on electron is 1.6 into 10 ^ minus 19 Kum okay and then your e so this is what you have to find out now you can bring it to the left hand side so 1.6 1.6 cancel 10^ - 16 divided by 10^ 19 - 19 and minus sign minus sign and I so the answer isus 10^ 3 I that is the direction and that is the value of your electric field so answer is minus 10 power 3 I option b super are you all clear electr charge electron charge- 1.6 into 10^-9 basic please very good very good super let's go forward so this is the next question let's see how many of you are able to solve it okay please try to solve it I need some break for my voice at least three chapters I will complete so one week I will complete three or four chapters at Max okay and uh next chapter will be Monday Solutions chemistry CH so that's the plan you have to answer this ex so please don't tell me you don't know don't tell me you don't know okay okay I want you to solve it I want you to try I'll give you one or two minutes you will get it I will teach you will learn you will get better okay great cbsc sheet release I'm also waiting for it so let's wait released it December elections soia so mostly next week we will get it I'll let you know very good very good Mr strange Lev level Mr strange is correct is Right Dave you are close da but that's wrong you are close da but that's wrong okay but you're close so I'll give it to you okay and har super haras mishar mishar that's not right that's not right so I will teach you please listen guys I will teach you wait okay it's okay okay please don't spam guys don't spam the chat okay unnecessarily you will get removed so let's not hurry electric field in Direction sir electric field is along xaxis sir it is along I Direction okay electri field so electric field X the the do you agree the flux through the Mal face Kila face pady face why sir electri field vular area Vector electric field perpendicular 90° 90° 90° so your area Vector is upwards electric field is along x- axis sir sir R 90° sir 90° now Ceta is M so flux is e a e do a e e a COS Theta cos Theta is so flux through the six faces very good do you all agree do you all agree are you guys crystal clear with whatever I have taught you until now so shall I move forward very good so length of the cube length l xal l so in the line xal L in the line xal M correct in the in the face in the face electric field is equal to a into mutai plus b correct and in the face electric field is equal to a into x + bow I want you to solve like this so electric field in the face L plus b number IAP Direction IAP now right side and a l plus b IAP Direction super clear so flux in the mod phas phas one pH two okay the flux in the mod phase is equal to e Vector do a vector sir area normal should be outward sir normal should be outward so it's going to be minus I cap it is going to be minus I a flux through the first phas will be e do A1 e vector. A1 Vector so e Vector b i so b i dot l s minus I cap so it's going to be minus BL s squ because you guys are awesome how x equal to Z origin origin orig x s I hope all of you are able to understand Masa so flux through the second phase Pap flux through the second phase is equal to electric field in the second phase into dot area of the area Vector of the second phase okay are you all clear everybody body happy electric field in the second in the second pH electric field is a l+ b i dot area Vector sir VOR second l square area length into side into side so square area L Square direction direction is outward normal so outward normal it will be along xaxis so it'll be I cap so area is area is l s i cap so flux flux in the second phase is equal to a l Cub plus b L Square sir first second equal I will get Al Cube so net flux M are you all clear everybody happy with this s so now the question is over it's a very good question it's a very very good question okay so yeah I hope you liked it explained it I have tried to explain everything find the charge okay I'm going to erase in five four 3 2 1 that's it so sir n charge I know I know that is there okay so let's solve it okay uh s IAP along xaxis IAP s jcap along Y axis J and alongis okay Direction okay soet charge we have to use net charge enclos by thebe so the the so in the cube surface the net flux is equal to charge enclosed by Epsilon is so net flux is equal to charge enclosed net charge Q enclosed now it is the net charge inside the cube Q enclosed now it is a net charge inside the cube divided by Epsilon KN so the charge enclosed is equal to Al Cub into Epsilon KN that's the answer are you all clear everybody everybody happy with this so shall I move forward so easy okay it's a nice question did you all like it so don't worry don't get scared okay great so let's go forward let's go to the next question yeah please try to solve ity m shini thank you so much I'm glad you're enjoying check let's see whether you are right or wrong PR 10 standard evening I will take for 11th or 12th okay follow so usually I take for 10th standard 5:30 6:00 6:30 on the time and 11 next live will be on Monday okay great electric of L 2 how many of you took AAL two please remember 2 is two okay will I took as as sir I knew please be careful don't do that that okay don't do that okay so electric dipole of length 2 cm is placed at an angle of 30° with an electric field so what is the torque tor vectoral okay he has given you all the values so let's go to magnitude so so we know equal p p p okay so now just plug in the values just plug in the values you're done soang 8 into 10^ minus 3 8 into 10us value what is p p it is nothing but it is 2 AQ okay what is the magnitude of charge so value p p p is 2 a q so 2 a so 2 a 2 cm into q q is the magn 5 e value 2 into 10^ 5 and sin Theta sin 30 is half is 30° so 30 is half I hope all of you can do it very good you guys are absolutely awesome so in the two the two cancel I want you to try and get it done so in the zero in the zero and in the two so 4 into 10us 3 is equal to Q into 10us 3 sorry 10^ 3 okay 4 into 10^ minus 3 is equal to Q into 10 the 3 q q 4 into 10us 6 Q it is 4 micro micro 10us 6 micro 10us 6 four micro is the answer s very good arut s boss you are right har level Sia very good afia super AIA Zia fantastic sham rant all of you are right how 2 by 100 Amir 2 cm am 2 cm 2 by 100 I hope all of you are a to follow div great so that's about it this is again a theoretical question please solve it what do you think is the answer if the net electric flux through a close surface is soos surface okay so in the closed surface I'm going to treat it as a Garian surface so Garian surface net flux is mut net flux through the Gan surface is zero that is equal to charge enclosed by Epsilon KN so this will imply that the enclosed charge is zero so the net charge enclosed is zero so did you all like it how many of you are clear can tell you cannot comment so don't go to option b or option C or option D I hope you so now coming to the next one so I will not help you I want you to do it come on guys don't give up I want everybody to get the answer so please question so don't get scared do it correctly okay so yeah try it try it try it everybody oh oh yeah really nice nice I know mura sir I have met him he's a very nice person okay everybody try to solve it it's a very nice question I want you to try and finish it off okay so positive it is along plus X Direction so XIs electric field positive positive X it is to the right and it is in the left so X positive it is to the right and x negative it is to the left okay very good very good great so I hope all of you have understood it so what do we have we have a right circular cylinder so we have a cylinder CM Center is at origin soer symmet are you all clear everybody happy superb superb very good we are going to find out the net outward flux through the cylinder so outward okay so everybody listen carefully I want you to answer one thing okay I want you to answer one important thing in the curved surface area what is the net flux can anybody let me know in the curved surface are through the curved surface area in the curved surface area in the circle okay so the flux is zero okay great great great so flux formula it is electric field dot your area Vector so electric field it is along x-axis okay so in the in the bottle cylinder electri okay the curved surface area Vector will be perpendicular everywhere the area Vector for the curved surface area will be perpendicular everywhere so perpendicular so the flux will be zero because because are vector electric field angle 90 so the flux in the r surface the curved surface area of the cylinder the flux in the Rend surface the curved surface area of the cylinder is M okay very good so the in the first surface is equal to electric field in the first surface into area in the first surface dot product flux is e do a so the first surface electric field electric field is-200 I correct do you all agree minus 200 N it is along X electric field in the first surface is along I negative x negative electric field is along negative Direction I area is outward Normal area is so surf so first surface so first surface which is a circle so the area will be outward normal so outward normal sir it is towards negative I it is towards negative I so area is also along netive I area value area value it is pi r s it's a circle so circle area Pi R square so it is going to be Pi into 5 by 100 whole Square so how much it will be Pi into 5 by 100 whole Square can you all try doing this are you all happy my dear boys and girls yes all of you are clear very good so the minus in the minus cancel so the flux will be positive it is outward so it's a positive flux okay so it is going to be how much it's going to be 200 uh into Pi into 25 by 100 into 1 by 100 5 Square 25 100 square 100 into 100 so 100 remove and 25 50 it become by2 so Pi by2 is the flux from the first surface super so first surface first surface it is positive Flux Of Pi by2 so that is the answer so it is going to be outward are you all ready of flux Newton M Square per Kum if you want to keep the units also correctly Okay so let's try to see the second surface so second surface easy so flux in the second surface is equal to electric field in the second surface do area and the second surface okay sorry second third so third third let me write Third only and second surface second surface because curved surface area it is no because curved surface area area vector and electric field Vector are perpendicular area vector and your electric field Vector are perpendicular so second surface it is zero okay now third surface what is electric field electric field sir right side positive xaxis sir positive xais electric field it is plus X so positive xaxis electric field it is plus X so plus 200 I cap electric field dot Vector in the in the surface theface theface vect it will be like this it will be perpendicular to the circle so perpendicular to the circle in Direction sir perpendicular to the circle outward normal is along x-axis Direction sir it is along x-axis Direction so it is going to be along IAP Direction and what is the area area value it is pi r² so again same thing Pi into Pi by 100 whole Square so calculate again you will get Pi by 2 Newton M Square per sir by by Sir total total total total flux net flux is equal to 5 1 + 5 2 + 53 so first surface Pi by 2 second surface mutai third surface again Pi by 2 so the net flux is equal to Pi so the net flux is equal to Pi did you all like it the net flux is equal to Pi are you all clear with this everybody happy very good super soge easy charge easy okay i'me this as the Gan surface so in the cylinder Gan surface so in the cylinder Gan surface now it is a closed figure so if is a close figure the net flux the net flux is equal to charge encl closed the net charge enclosed by Epsilon so net flux that is equal to the net charge enclosed by Epsilon KN so the net charge enclosed is equal to Epsilon KN into Pi so Pi Epsilon is the answer you will get full Mark okay good question right please study this so you have to study all of this so you have to study this don't don't let go of this it's a it's a good question okay so qal epon P value 3.14 approximately and Epsilon 8.854 into 10^ minus 12 so charge value enclosed charge so the net charge enclosed is equal to Pi into Epsilon that is your net charge enclosed you will get the answer that is the right answer it's it's over okay cool so shall I move forward shall we move to the next question are you all ready direction is already assumed here the electric field is in the negative X Direction in Negative X so left hand side of r it will be to the left right hand side of it be to the right very good let's go to which is over here will find it hard so don't worry we will learn it we will learn it okay thousand Reasons right lock okay so yeah please try this this let me know okay it's a nice question I want all of you to try it [Laughter] out great very good Z V level Z Super Z you right okay lakmi super lakmi level so pangla two small identical dipoles a and CD are are kept as shown in the figure and they have an angle of 120° between each other so you need to find the net dipole moment okay so you have to overcome that it's it's very easy okay like this so AB dipole it is going to be p in a direction J cap correct do you all agree and number CD so CD in the direction equal vectors resultant resultant is equal to square root of p² + p² + 2 p into P into cos 120 how many of you understood that so the resultant is also P so the net dipole moment is also P can you believe it the net dipole moment is also P isn't that crazy how many of you are able to follow in a dipole moment or vector identical magude magnitude same so resultant is what result you would have studied this formula resultant of a vector and + b² + 2 a COS Theta so resultant of two vectors which have an angle Theta in between them so Vector p b Vector VOR so you can use this and you can get the result so I will teach you one more method law vector addition vect Okay so let's do it theole it is making an angle 30° with xaxis in the CD dipole in the CD dipole it is making an angle 30° with the x-axis are you all able to follow that very good super so I'm going to resolve that so CD Dio which is magnitude p and it makes angle 30° with your xaxis sove CD dipole it will be P cos 30 P cos 30 I plus P sin30 J cap minus JC do you all agree so CD dipole value how many of you are able to understand so CD dipole okay are you all clear moment asking Dio moment of the arrangement so what is the net dipole moment so you have to add them you have to add them Vector okay P horizontal P sin 30 vertical down so P cos 30 I cap cos 30 root3 by2 and P sin 30 J cap downward Direction minus J cap okay so I'm going to add both of them because I want the resultant I want to add both of them and see it and J p p by2 so P minus p by2 p by2 sir so the resultant so this is the resultant over what is the magnitude magnitude is square of I component plus square of J component added together under root so magnitude is p so the resultant magnitude is p only the resultant mag ude is p only level M did you all enjoy it are you all happy my dear boys and my dear girls 120 minus 90 how okay very good very good super they what is theole moment of the arrangement you have to tell direction also you have to teach the direction also okay abish electrostatic potential and capacitance please wait okay so it is3 I P by2 JC okay p root3 by2 i plus P by2 JC value root3 by2 i j p by2 so result is going to be like this correct either JC value p by2 i positive I positive JC so JC value p by2 and I value pk3 by2 in the angle tan Theta tan Theta is equal to opposite by adjacent so Theta 30° so the resultant is p magnitude it makes an angle of 30 with the Positive XIs nobody can stop you okay did you all like it was it V level did you all enjoy it so answer is it is going to have a magnitude of p and what is the direction dipole is a vector right you need to give me direction also so what is the direction the direction is 30° with the Positive xaxis okay great magnitude and direction of torque so so your electric field is along X Direction okay so torque it is p cross e okay so let let's try to do that it's it's very simple only it's not great okay so so p p this is one p and this is other P okay so torque net torque is equal to P cross e for the first one and P cross e for the second one so first one what is the angle between p and E sir first one e along xaxis sir so angle is 90° correct AB the angle between p and E electric field electric field is along xaxis so along xaxis electric field so e p angle 90° so it will be P sin 90 plus second one second one sir p x angle between them angle between them is De angle between them is De P cross e p cross e is downward into it is into the paper it is minus K cap because of ab ab p p Vector cross e Vector P Vector cross e Vector is into the board into the board now minus K cap Plus value P cross e p e sin 30 sin 30 and it will be what sin 30 will be K cap Direction P cross e so P cross e will be outward so it will be K capap Direction so the net torque the net torque is going to be uh minus P by 2 K cap so it's going to be P by 2 K cap inside so that is how you solve it not like okay are you all clear yes did you guys like it s good question please great okay bang I'm happy that you all understood it now let's move forward so let's go to the next one great so please try it out everybody try to solve it so I will repeat the same question so please sir M zis okay so xaxis Direction Vector J yis unit vector and k z axis Direction unit Vector so positive X unit Vector positive y unit vector and positive Vector okay great so oia has answered so Pang number mic micro and R they are separated by a distance of 30 cm they are separated by a distance of 30 cm so where will the electric field between the charges be zero it will bether from the bigger charge number net electric field zero so let's take it as X from the 1um one micr charge 3us x sir3 30 by 100 so that will be3 M okay always do it in meters always do it in meters So Sol so electric field at P because of 1 micrum charge is equal to KQ by r² and it will be IAP Direction and electric field at P because of 4 micrum charge will be k q by r² and it will be minus IAP Direction so the net electric field at P the net electric field at p is zero so electric field electric field at P because of + one micrum plus electric field at P because of plus4 micr they should add up to give zero so so Z I will get K into 1 micr by x² minus K into 4 microc by3 - x² is equal to I I I I so I should get zero add I will get this minus this I minus I so subtract I'll get equal to zero so if I take this to the right side it will become like this so k k cancel micrum micrum cancel so I'll have 3 - x by x² is equal to 4 3- X by xal 2 so 3 is - x is = 2x so 3 is = to 3x so X is equal to 0.1 so the answer is it is 0.1 M from the + one micrum charge so answer is X = 0.1 M from + 1 micrum charge did you all like it was it good did you all enjoy it CLE are you guys happy please let me know my dear boys and girls don't give up very good so now let's move forward so guys are you all having fun squ or fire I want you guys to be happy I want you guys to be energetic okay don't give up guys it's okay okay so electric field along X direction is minus DV by DX so the form electric field along X direction is the gradient negative gradient of your voltage across X ch potential potential X okay great okay so I hope all of you are clear so now we will study in the next chapter that your electric field along a certain direction can be found out in this manner it is the negative gradient of your potential it is a negative gradient of your potential beus 10 correct either if you differentiate with respect to x what is DV by DX DV by DX is equal 10 and electric field is minus of DV by DX so minus of DV by DX will be minus 10 so this is the magnitude of your electric field along X Direction so electric field will be like this okay so electric field Vector wise it will be minus 10 I that's the answer so Total Electric flux through the cube total electric flux through the cube is answered very good z z you absolutely right good job very good level B you're right so electri field electric field is along X electric field is along Negative X and it's a constant so you have a uniform electric field towards Negative X Direction which is not changing so total flux zero so zero surface okay Mal Square Vector isend to the electric field area Vector will be outward Normal area Vector will be outward normal so area Vector is perpendicular to your electric field so flux is M Okay so is also so that will be perpendicular to your electric field Electric fieldi so electri electric field is like this IAP Direction so the first surface the second surface so electric field in the first surface is - 10 I area in the first surface is area into minus I so flux flux in the first surface is e do a which will be how much which will be 10 a so flux through the second surface calculate it is equal to area in the uh electric field in the second surface dot area in the second surface electric field in the second surface is again minus 10 I area in the second surface is equal to a i second surface area is outward normal outward normal will be along X Direction so R surface area Vector will be IAP Direction surface area Vector will be IAP Direction so e do a flux will be e do a minus 10 a so net flux net flux is equal to pi1 + 52 uh which is equal beautiful right so good so good level okay so are you are happy are you all having fun are you all able to enjoy are you all able to learn whatever I've have covered here yeah good super okay next one please solve it okay so electric flux flux is equal to e vector. a vector okay so uh e Vector do a vector so Square plane so square area square area is 10 by 100 into 10 by 100 square area side into side so the area is 1 by 100 okay so area of magnitude is 1 by 100 so flux is equal to e a COS Theta okay so e e 200 area is 1 by 100 and cos Theta is how much cos Theta is cos 30 cos it is cos 60 cos 60 is 1 by two so the answer is 1 Newton M Square per Kum so answer so you have a positive flux okay great so that's the answer so electric field you took the angle 60 squ square ELC electric field so electric field square square is the between our area vector and the electric field so area Vector area Vector is normal to the surface of the square area Vector is always outward Normal area Vector is always outward normal so this will be 60° I put 60° now do you understand so this is your area Vector are you guys Happy okay so let's go to the next question so please manage with this so we almost done we're almost done okay so I hope you're all excited so shall we solve itam years 2014 2013 I wanted to cover everything so please great so what do we have we have three charges on three sides three vertices of an equilateral triangle okay so you have charge at a which is 10 micrum charge at B which is 5 micrum and char at C which is - 5 micrum okay so we have to find the resultant Force experienced by the charge at a so B the force will be like this correct so the force on a because of B is equal to K q1 Q2 by R squ what is the side of the equal triangle 0.1 okay are you all clear uh I hope you guys are happy with this okay so Cate if you want can you calculate it so force on a because of B is what K 9 into 10 ^ 9 so 9 into 10 ^ 9 10 into 5 is 50 micrum micrum will become 10^ - 12 by 0.1 Square so how much will that be 0.1 Square it will become 100 0.1 Square Mal 100 okay so it'll be 9 into 5 45 10 into 100 will become th000 so th000 cancel so it is 45 Newtons correct 45 I think it's correct so it is 45 Newtons in this direction direction okay so C on a c on a will also be attractive this will be repulsive and positive positive will repel negative positive will attract so so force on a because of C magnitude magnitude please magnitude so magnitude again K into q1 into Q2 by r² Sir Sir magnitude magnitude k k 10 micr 10 micr 5 511 Direction fix negative charge will attract this so it will be towards that I'm not I'm going to use geometry so if they if they're equal the resultant is also 45 Mev did you all like it the resultant is also 45 how many of you enjoyed it 120 root of 45 squ + 45 squ + 2 into 45 into 45 into cos 120 cos 120 cos 120 it is minus half cos 120 is minus half so solve resultant will also be 45 so 45 equ equ the will be symmetric so 120 6 so the resultant will make an angle of 60° with your AC so sir is 60 it is this is also 60 sir equ triangle so equal in the resultant is parall to BC so the resultant is parall to BC and it is of magnitude 45 that is full marks I have given you full Mark okay such a good question a level M SC we okay are you all happy so shall we all move forward great so yeah I hope you enjoyed it it's a simple question next one this is also easy do it fast okay okay what is the answer so in the easy I hope all of you will do it the following figure shows a surface so Gan surface so Gan surface encloses minus 2 Q charge plus Q charge is kept outside calculate the the net outward or inward flux so net flux for aian surface which is a closed surface it is equal to enclosed charge by Epsilon KN so net flux answer so net flux is negative so negative it is inward flux it is Inver flux because it is negative sir in the plus Q field lines they will pass through thean surface are you all happy so for the net flux so net flux is only because of whatever charge is present inside the Garian surface face okay that's it next one very very doable question again a charge Q is enclosed in a cube what is a flux associated with one of its faces very good so okay so the net flux net flux is equal to charge enclosed by Epsilon KN it's a equal distribute so each pH will have q by 6 Epsilon kn okay are you all clear I hope you guys enjoyed it so net flux coming outside the cube will be Q by Epsilon not given that Q is a positive charge and that net flux is coming through all the six phases so each phase will be Q by 6 Epsilon that's the right answer so I hope this session was useful in the standard classmates juniors seniors okay and if you want to test yourself J if you want to yourself you can write our mvat exam MV exam registration in link it's a free of C exam it is full portion for 12th standard it is full portion for 12th standard okay great so that's about it do join theam group and finally if you are new to the Chanel and I will see you in the next one okay so take care boys and girls love you all see you in the see you in the next one okay 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