[Music] okay good day so this is our last topic for general physics tool which is electric flux and gas law so our learning competency is to be discussed for this period are first calculate electric flux and then second is to use gauss law to infer electric field due to uniformly distributed charges on wires spheres and large plates okay but before we proceed with our main topic let us recall what we have learned on our previous discussion okay so um regarding electric field lines okay as we know electric field is the space around an electric charge so electric field lines may differ different depending on what is our point charge is for positive charge okay the electric field line is um directed outward the the point charge um on the other hand if negative pneumonium point charge not any electric field line is directed towards the point charge okay so when this charges interact okay with one another okay so they create this um representation of attraction when two um opposite charges move closer to one another they attract and when we have two light charges they repel each other okay so this is through the calculations of the electric field magnitude using the equation e is equals to force per given value of test charge okay so based by deriving another equation for electric field from coulomb's law we have another equation which is e is equals to k multiplied by the value of our point charge over the distance squared okay so now um let's move to our main topic which is electric flux okay as we all know um each charge particle okay distributed its lines or it distributed its force due to or with the use of electric field lines okay so this okay for example we have um a hollow box okay and we put an individual charge or a single charge inside a hello hollow box and then um on the second diagram we have the same hollow box same size but um we tended to put the two charges inside it okay as we can see on the right part of our diagram this is this term or this phenomenon is what we call electric flux okay it is the change or it is the rate of a flow of electric field through a given surface okay so you spread out no adding electric field okay illustrated by lines here in our diagram is what we call the electric flux so electric flux um we defined it as um the electric field through a surface okay given the equation fee okay phi is the denotation of electric flux is equals to the dot product of electric field and the surface area okay the surface area vector of that specific object okay and then given to expand this that product of two vectors electric flux v is equal to the electric field multiplied by the area vector or surface area cosine theta the theta here is the angle between our electric field and our area vector okay in some notes or in some books this area vector can be denoted as um the normal vector see area vector is always perpendicular to the surface of our object okay it can be a box a balloon or a cylindrical a cylindrical surface okay so the um unit used for electric flux is equal to newton meters squared per coulomb okay so it is denoted by a greek letter phi okay so um the electric flux can be calculated in different orientations okay let us um focus on this first orientation electric calculate the flux of the electric field e okay given this e through the surface area a okay this is our surface area in each of these three cases shown so this is the first um case okay so the area vector here as we know okay it can be the normal vector and can be the area vector can all it also it is always perpendicular with our surface so the direction of our area vector is towards this direction okay the the electric field vector the direction of it okay the angle between the angle between our electric field vector and our area vector is equals to zero okay case zero degree if we would be applying the equation for electric flux phi is equals to e a cosine theta okay so zero degree cosine zero degree is equals to one so the value on this kind of orientation of our electric flux is cosine zero is one is at the maximum so we have to multiply um the magnitude of the electric field to our surface area vector okay since i'm cosine zero is equals to one so most probably in this kind of orientation the magnitude of our electric flux would be actually maximum okay this um let's now let's move forward with our second type of orientation so um comparing uh from our first diagrams okay so same again you calculate that you most probably value of our electric flux by analyzing this kind of orientation but first [Music] a direction of our area vector so you are your vector again parallel with our surface direction okay our field vector or yes our field vector is going through the part of this part so this is the direction of our electric field so okay angle theta the angle between electric field vector and area vector so what is supposedly the direction between them so it is electric field vector [Music] would be 90 degrees okay so applying the equation for um electric flux ea cosine theta so our theta we would be using 90 degrees cosine 90 degrees is equals to negative one cosine 90 degrees is equals to zero cosine 90 yes cosine 90 is equals to zero so the calculated electric flux for this kind of orientation is equals to zero ea multiplied by zero okay so if we have this kind of um diagram where our surface is perpendicular with our field or don't there is no field um passing through okay a specific surface surface so most probably the amount of electric flux okay on that surface would be equals to zero what is our definition of electric flux it is the flow of electric field that penetrates a surface okay and then last this um orientation okay so we have here our third orientation where our surface is slightly slanted okay um but in comparison with our first diagram so calculate the electric flux of the electric field e through a surface a and in each of the three cases show okay so um you adding electric uh this is our uh electric field vector the first line here and this is the perpendicular line to the surface is our um area vector so most probably the distance or the angle between them is less than 90 degrees okay so compute and based from our equation e a cosine less than 90 degrees okay it would have a value of um uh at the middle for example i think theta here is 60 okay 60 degrees ea cosine 60 by cosine 60 is one half so our value would be ea okay times one half so how about 16 angle pa what if our angle theta is greater than 90 degrees so upper greater than 90 degrees cosine greater than 90 degrees is um equal to negative smaller of magnitude so most probably if negative young electric flux not in okay if negative and electric flux that in the electric field the e field is is coming from okay coming from the outside to inside so what does this mean okay meaning field is coming from vice versa coming from the inside to outside so meaning your electric field and orientations of electric flux first kapaka okay kapaka cosine meaning the electric field vector and the area vector is at the same direction okay the angle between them is zero okay cosine zero degrees is equals to one so the value of the electric flux is at the maximum okay so beaten amanda cosine okay your ea your electric field and area vector are perpendicular with one another perpendicular okay the angle between them is 90 degrees cosine 90 is equals to zero so meaning the value of our electric flux is equals to zero meaning the there is no field vector that penetrates a surface and then lastly if we have a value of or the angle between ea electric field vector and area vector is less than 90 so your magicking um value is a positive flux okay flux density is positive meaning um it comes from inside to outside on the other hand okay on the other hand if we have an angle greater than okay greater than 90 degrees electric flux the value of it would be negative okay so the field line comes from outside to inside okay so that is um how we calculate electric flux depending of the orientation of our surface okay so what if our surface is curved or um the field varies with position electric field surface okay based from our discussion we know that um electric flux is equals to the dot product of electric field vector in area vector so parama kohanaten electric flux on the varying position we have to divide the surface area into small regions surface okay um differential of our air yeah or in terms va okay so the flux through that that small part that we get into the surf the region of a surface is equal to um using that differential area we can calculate the flux d flux okay or d phi is equals to e the electric field vector multiplied by the differential of of an area cosine theta or um electric flux differential of electric flux is equals to e that product of the a okay so to obtain okay to obtain the total flux density we need to integrate um over the surface area okay in terms of v so since we are getting phi integrate both sides okay d phi is equals to integrations of e dot product of d a so this is how we get okay the value of the flux density if we have a varying um field or position of a field in terms of a curved surface direct algebra we need to get a small part of that surface and then integrate it for us to get the total flux density okay this is for open spaces or open surfaces what do we mean by open surfaces for example we have a um cup or we have a coffee mug okay so um uh uh looking at it and you can also see the smoke coming out from a hot a hot coffee okay that is an open area so of our open surface from um created by the coffee okay as a bathroom evaporates okay so um okay so this closed surface is also called as the gaussian surface this is symmetrical okay meaning all of the parts of this surface um is or do have the same sizes okay so perform on a closed surface the flux is considered to be positive for filled lines that leave the okay the enclosed volume so hapaga and field lines again same as uh the open surfaces uh it's coming from the inside to the outside um our flux density is positive our flux density is negative if the filled line snapping is entering okay is entering our enclosed volume or um the flux or the filled magnitude is coming from the outside to inside okay so same language equation for open surfaces and closed surfaces or gaussian surfaces we we just need to indicate this um symbol okay so in integration symbol to represent that we are getting okay the flux density okay the electric flux density and the electric flux casimiro and the entire magnetic flux okay so the the flux density on a closed surface okay now nabangit nathan young gaussian surface okay this gaussian surface is derived from ghost law or gauss so this law states that um or this law is an expression of the general relationship between net electric flocks through a closed surface and the charge enclosed by the uh surface okay so pharmaceutical in the hand let's take a look on this diagram so maritime sphere okay this is um example of a gaussian at the middle part of it my time is on charge a positive discharge in this case they are here is um the distance or the reviews um with respect of on the middle of uh the spear so you know they use yeah getting the part of this region va or this area vector would be um perpendicularly originally coming from the middle outward perpendicular to the surface it has the same direction as our electric field vector so d a is always parallel um if we are talking about symmetrical closed surfaces according to gas law the our um differential of the area vector is parallel to our electric field vector at any point of ours spear okay so the angle between them angle theta between them is equals to zero parallel direction okay so if we are going to use that with our equation of electric flux v so e a cosine theta since um v a is parallel to e so e a electric flux v e a cosine theta or cosine zero is equals to one so phi is equals to e a okay so since um our gaussian surface here is a sphere in order for us to get the surface area of a sphere we um use this equation area of sphere is equals to four pi okay r squared so prediction is flux e four pi r squared tapos the electric field vector from our previous discussion is equals to um coulomb's constant multiplied by uh the magnitude of a point charge over r squared okay so padding is substituted okay k q a 4 pi [Music] r squared all over r squared as we all know columns constant is equals to one over four pi okay epsilon naught okay so it's a substitute so q 4 pi r squared over 4 pi epsilon naught okay r squared so my my meron a time my kahn cell so you cancel out four pi cancel out r squared somaliland equation of electric flux okay which is q over epsilon naught okay what is epsilon naught epsilon naught is um what we call as the permittivity of the permittivity of uh free space so it is also equal to the value of 8.85 so this is a constant value 8.85 times 10 raised to negative 12 newton meter squared okay so if we have that is for a single charge okay um using gauss law calculating um electric flux density on the gaussian surface or at a closed surface with a single charge we use this okay we use this equation um gaussian surface so based from our equation phi is equals to q over epsilon naught so we have to find the mag the summation or the value of all charges that can be found on an enclosed system so our equation would be fee is equals to summation of the charges in an enclosed surface all over epsilon naught okay so that is how we get um the value of the flux density with more charges on a closed surface or the total electric flux so in summary so dos law can be mathematically defined using this equation fee is equals to summation of q on an enclosed system or total magnitude of charges in a gaussian surfaces over or divided by the permittivity of free space which is equals to 8.85 times 10 raised to 12 negative 12 newton newton newton meter squared okay so as we define gauss law this law according to carl frederic friedrich gause this law is or states that the electric flux going through a closed surface is the sum of all charges inside the closed surface divided by the permittivity of free space okay so when we are going to calculate the total magnitude of um or the total density of electric flux going or that penetrates a gaussian surface which is also known as closed surface we need to find the summation of all charges and divide it with the value of permittivity of free space okay so that is um all for this lecture before i ended up my discussion let me share this quote to you um this question what is a soul okay it is like an electricity we don't really know what it is but it is a force that can light up according to real trials okay so that would be all for this lecture i hope you have learned something from electrostatics okay so we have learned electric forces electric field and now electric flux and gasoline