00:01
Consider point charge q is placed at this point and we draw a gaussian surface in the form of a sphere of radius r, say means around point charge.
00:26
So gaussian surface is a sphere of radius r.
00:30
On the surface of this sphere, we consider a small area.
00:37
D a right and this is an area perpendicular to to the surface so this vector means corresponding to this area will be in this direction right and also we know the electric field due to a point charge always points away from the charge so we can if we draw the means electric field due to this charge it will be rarely pointing away from the charge.
01:14
So this will be the direct means direction of the electric field and we know that the total flux passing through the gaussian surface.
01:26
This is our gaussian goss surface, right? the gossian surface is a surface of a sphere.
01:43
So total flux, total flux passing through the gaussian surface d5 will be equal to, electric field dot product with the vector area.
02:01
So it is going to be the electric field vector area multiplied by cost of the angle between them.
02:11
Now here e and da are parallel.
02:15
So theta is equal to zero.
02:19
So, cos theta is 1, right? so the total flux passing through the gaussian surface will be simply equal to e multiplied by da, right? their magnitude.
02:39
So this is in magnitude.
02:42
So now total flux will be, is the, it means electric field intensity integration of da, and da is a part of an area of.
02:54
Sphere r so total total area will be e into total total total means area of the sphere is 4 pi r squared is equal is the value of the total flux now as per the gauces law we know that total flux passing through a gaussian surface total flux passing through a gaussian surface is equal to charge in close in the surface divided by epsilon which is a constant so as per the goss's law phi should be equal to q upon epsilon we put this value of phi here so q upon epsilon is equal to e into 4 pi r square if we solve we will get the expression, the magnitude of the electric field intensity due to a point charge q is 1 upon 4 pi epsilon q by r square...