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Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields

A solid spherical region having a spherical cavity whose diameter ‘R’ is equal to the radius of the spherical region, has a total charge ‘Q’. Find the electric field and potential at a point P as shown.

1 Answer

Charge density $p=\large\frac{Q}{\Large\frac{4}{3} \pi [R^3 =(R/2)^3]}=\frac{6Q}{7 \pi R^3}$
Using the superposition principle,
$V = \large\frac{P (4/3 \pi R^3)}{4 \pi \in_0 x} -\frac{P [4/3 \pi (R/2)^3]}{4 \pi \in_0 (x - R/2)}$
$V= \large\frac{PR^3}{3 \in_0} \bigg[ \large\frac{7x-4R}{4 x (2x-R)}\bigg]$
$E= \large\frac{PR^3}{3 \in_0} \bigg[ \large\frac{1}{x^2} - \frac{1}{R (2x-R)^2}\bigg]$
answered Jun 2, 2014 by meena.p
 

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