Question

A point charge of -2.10 �C is located in the center of a spherical cavity of radius 6.54 cm inside an insulating spherical charged solid. The charge density in the solid is 7.36 x 10-4 C/m3. Calculate the electric field inside the solid at a distance of 9.48 cm from the center of the cavity. (magnitude and direction)

Answer 1

Answer:

$-3.4\times 10^5 N/C$

Explanation:

We are given that

$q_0=-2.1\mu C=-2.1\times 10^{-6} C$

$1\mu C=10^{-6} C$

$r_1=6.5 cm=6.54\times 10^{-2} m$

$1 m=100 cm$

$r_2=9.48 cm=9.48\times 10^{-2} m$

$\rho=7.36\times 10^{-4} C/m^3$

We have to find the electric field inside the solid at distance of 9.48 cm from the center of the cavity.

Volumetric charge density,$\rho=\frac{Q}{V}$

Charge on spherical solid=$Q=V\rho=\frac{4}{3}\pi(r^3_2-r^3_1)\rho$

$Q=\frac{4}{3}\pi((9.48\times 10^{-2})^3-(6.54\times 10^{-2})^3)\times 7.36\times 10^{-4}=1.76\times 10^{-6} C$

Electric field =Electric field due to spherical charge solid +electric field due to charge at center

$E=\frac{KQ}{r^2}+\frac{Kq_0}{r^2}=\frac{k}{r^2}(Q+q_0)$

Where $k=9\times 10^9$

$E=\frac{9\times 10^9\times}{(9.48\times 10^{-2})^2}(1.76\times 10^{-6}-2.1\times 10^{-6})=-3.4\times 10^5 N/C$

Where  negative sign indicates that the direction of electric field is inward.