Question

A smaller sphere is removed from a larger sphere.The sphere is an insulator of charge "q".
We have to find the the electric field due to remaining portion at point A.

Answer 1

@yrelhan4 @electrokid

Answer 2

this is a repeated problem..

Answer 3

find \(\sigma\)=charge per unit area
when the smaller circel is removed, the charge on the Area gets removed
then the same old

Answer 4

Can we solve this with negative mass concept?

Answer 5

I kept back the removed thing..i want to solve this way

Answer 6

http://jkwiens.com/2007/10/24/answer-electric-field-of-a-nonconducting-sphere-with-a-spherical-cavity/

Answer 7

let know if you got lost in it

Answer 8

That is for electric field in the cavity,i want remaining thing

Answer 9

@JFraser

Answer 10

for outside, it does not matter
the sphere now acts like a point charge but with a smaller total charge!!

Answer 11

\[
\sigma={q\over4\pi 4R^2}={q\over16\pi R^2}\\
Q_{\rm left}=\sigma \times\left(16\pi R^2-4\pi R^2\right)={q\over16\pi R^2}\times12\pi R^2\\
Q_{\rm left}={3q\over 4}\\
E_A=K{Q_{\rm left}\over d_A^2}
\]

Answer 12

Answer will show that electric field is uniform in the cavity.
You can solve by superposition of positive and negative charges, no problem, since laws of electromagnetism are all linear.

Answer 13

@Vincent-Lyon.Fr not In the cavity..
outside the system at a point "A"