1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

022-62211530

Mon to Sat - 11 AM to 8 PM

Using Gauss's law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.

Asked by Topperlearning User 17th April 2015, 2:01 PM

Consider a spherical Gaussian surface of radius r (›R), concentric with given shell. If is electric field outside the shell, then by symmetry, electric field strength has

same magnitude on the Gaussian surface and is directed radially outward. Also the directions of normal at each point is radially outward, so angle between and is zero at each point. Hence, electric flux through Gaussian surface =

Now, Gaussian surface is outside the given charged shell, so charge enclosed by the Gaussian surface is Q.

Hence, by Gauss's theorem

Thus, electric field outside a charged thin spherical shell is same as if the whole charge Q is concentrated at the centre.

Graphically,

For r ‹ R, there is no strength of electric field inside a charged spherical shell.

For r › R, electric field outside a charged thin spherical shell is same as if the whole charge Q is concentrated at the centre.

Answered by Expert 17th April 2015, 4:01 PM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer 10/10