Answer to Question #104210 in Electricity and Magnetism for Akshay

Question

A non-conducting sphere of radius 1.0 m carrying net positive charge18nC is enclosed by a concentric non-conducting thin spherical shell of radius 2.0 m carrying net negative charge 20nC. Determine the electric fields at the distances of 0.5 m, 1.5 m, and 2.5 m, respectively, from the centre of the sphere.

Answer 1

(1)$r=0.5m$

According to the Gauss's law

$\int {E\cdot dA}=\frac{1}{\epsilon_0}\sum q_i$

$E\cdot 4\pi r^2=\frac{1}{\epsilon_0}\frac{q_1}{4/3\cdot\pi \cdot R^3_1}\cdot 4/3 \cdot\pi \cdot r^3\to E=\frac{q_1}{4\pi \epsilon_0R^3_1}\cdot r=$

$=\frac{18\cdot10^{9}}{4\cdot 3.14\cdot 8.85\cdot 10^{-12}\cdot 1^3}\cdot 0.5\approx162V/m$ Answer

(2)$r=1.5m$

$\int {E\cdot dA}=\frac{1}{\epsilon_0}\sum q_i$

$E\cdot 4\pi r^2=\frac{1}{\epsilon_0}(q_1)\to E=\frac{q_1}{4\pi \epsilon_0r^2}=$

$=\frac{18\cdot10^{-9}}{4\cdot 3.14\cdot 8.85\cdot10^{-12}\cdot 1.5^2}\approx72 V/m$ Answer

(3) $r=2.5m$

$\int {E\cdot dA}=\frac{1}{\epsilon_0}\sum q_i$

$E\cdot 4\pi r^2=\frac{1}{\epsilon_0}(q_1-q_2)\to E=\frac{q_1-q_2}{4\pi \epsilon_0r^2}=$

$=\frac{(18-20)\cdot10^{-9}}{4\cdot 3.14\cdot 8.85\cdot10^{-12}\cdot 2.5^2}\approx-1.31 V/m$ Answer