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

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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.

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

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

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