Question

A thin spherical shell of Metal has a radius of .25m and carriers a charge. 2*10^-6C. calculate the electrostatic intensity at a point. a) inside a shell. b) on the shell and 3m from the center of the shell. ​

Answer 1

EXPLANATION.

Thin spherical shell of metal has a radius

of 25m.

$\sf \: charge \: carries \: = 2 \times {10}^{6}c.$

To calculate the electrostatic intensity at a

point.

(1) = inside a shell.

The charge intensity inside a shell is = 0.

(2) = on the shell.

$\sf \: E = \dfrac{1}{4\pi \epsilon _{0} } \times \dfrac{q}{ {r}^{2} } \\ \\ \sf \: E \: = \frac{kq}{ {r}^{2} } \\ \\ \sf \:E \: = \frac{9 \times {10}^{9} \times 2 \times {10}^{6} }{ {25}^{2} } \\ \\ \sf \: E \: = \frac{18 \times {10}^{15} }{625} \\ \\ \sf \: E = 0.0288 \times 10 {}^{15} \frac{n}{c}$

(3) = 3 m from the center of shell.

$\sf \: E \: = \frac{kq}{ {r}^{2} } \\ \\ \sf \: E \: = \frac{9 \times 10 {}^{9} \times 2 \times {10}^{6} }{ {3}^{2} } \\ \\ \sf \: E \: = \frac{18 \times {10}^{15} }{9} \\ \\ \sf \: E \: = 2 \times {10}^{15} \frac{n}{c}$

Answer 2

EXPLANATION.

Thin spherical shell of metal has a radius

of 25m.

$\sf \: charge \: carries \: = 2 \times {10}^{6}c.$

To calculate the electrostatic intensity at a

point.

(1) = inside a shell.

The charge intensity inside a shell is = 0.

(2) = on the shell.

$\sf \: E = \dfrac{1}{4\pi \epsilon _{0} } \times \dfrac{q}{ {r}^{2} } \\ \\ \sf \: E \: = \frac{kq}{ {r}^{2} } \\ \\ \sf \:E \: = \frac{9 \times {10}^{9} \times 2 \times {10}^{6} }{ {25}^{2} } \\ \\ \sf \: E \: = \frac{18 \times {10}^{15} }{625} \\ \\ \sf \: E = 0.0288 \times 10 {}^{15} \frac{n}{c}$

(3) = 3 m from the center of shell.

$\sf \: E \: = \frac{kq}{ {r}^{2} } \\ \\ \sf \: E \: = \frac{9 \times 10 {}^{9} \times 2 \times {10}^{6} }{ {3}^{2} } \\ \\ \sf \: E \: = \frac{18 \times {10}^{15} }{9} \\ \\ \sf \: E \: = 2 \times {10}^{15} \frac{n}{c}$