CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

(i) At a point P outside the spherical shell :
Let E be the electric field at P. The electric field through an elemental surface ds is
dϕ=E.ds
Total electric flux through the gaussian surface is
ϕ=SEds=ESds=E×4πr2
ϕ=qϵ0
E×4πr2=qϵ0
E=14πϵ0.qr2 (for r>R)
(ii) At a point P inside the spherical shell:
In this case charge enclosed =0
E×4πr2=0
E=0 (for r<R)

560939_522000_ans_ff62bea55b374c0ab6c011fcbea20a33.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Gauss' Law Application - Infinite Sheet
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon