Question

The electric field intensity at p and q in the shown arrangement are in the ratio with answer

Answer 1

we know, electric field intensity inside a conducting spherical shell is 0.

electric field intensity outside the conducting spherical shell is given by , $E=\frac{Q}{4\pi\epsilon_0r^2}$
where Q is charge and r is the separation between charge and point of observation.

see figure, charge inside the sell is q and outside the shell is 3q.

thus, at point P ,
electric field at point P due to inner shell is
$E=\frac{q}{4\pi\epsilon_0r^2}$
due to outer shell , electric field intensity = 0 [ because q is inside the outer shell ]

at point Q,
electric field intensity , E'= $\frac{q}{4\pi\epsilon_0(2r)^2}\{\textbf{inner shell}\}+\frac{3q}{4\pi\epsilon_0(2r)^2}\{\textbf{outer shell}\}$

= $\frac{q}{4\pi\epsilon_0r^2}$

thus, ratio of electric field intensity at P and Q = 1 : 1

Answer 2

Answer:

And given in image below