00:01
Hi, here in this given problem for the cylindrical capacitor, inner radius and outer radius are given.
00:19
Inner radius r, smaller, 5mm, outer radius r, capital r, that is 15mm.
00:34
Potential difference between these two cylinders, coaxial cylinders, that will be equal to potential of outer cylinder, which is 100 volt, and that of inner one which is 0 volts.
00:46
Is 100 volt only.
00:50
So electric field should be given by gradient of potential db by dr in magnitude only, gradient of potential 100, dr gap between the two cylinders, means it will be equal to difference in the radius of the two cylinders.
01:17
But no, we have to find electric field at point which is 10mm so dr will be 10mm minus 5mm no but no need to find electry field with this method we will use gossis serum to find electry field at a point midway between the inner cylinder inner cylinder whose radius is 5 millimeter and outer cylinder whose radius is 15 millimeter and we have to find electry field at a point midway between them at a distance 10 millimeter this is so we will draw a gaussian surface passing through this midpoint in order to find electric field but before that we should have a knowledge about the charge and closed within this gaussian surface for which we should have the value of capacitance of this capacitor cylindrical capacitor this is given as 2 pi epsilon not l is the length of the cylinder divided by natural logarithm of r by r.
02:41
So charge over the cylinder will be given by q is equal to c into v.
02:46
For c2 pi epsilon not l into b divided by natural logarithm of r by r.
02:53
Now plugging in the known values here, we will be able to find this charge.
03:00
So for the time being we keep it as it is only.
03:05
Now using means to find electric field we use gossus theorem here which says surface integral of electric field at the desired point is equal to 1 by epsilon not times the charge and closed as there is a medium between two cylinders whose permittivity relative permittivity is epsilon r so for epsilon not this is epsilon not into epsilon r and for this surface integral of electrical of electrical will be given by elective field multiplied by area of the cylinder, which is given as 2 pi r dash into l.
04:05
R dash is the distance of observation point from the axis of the cylinder is equal to q...