Question

two large thin metal plates are parallel and close to each other. On their inner faces the plates have surface charge densities of opposite signs and of magnitude 17.0×10^-22cm^2. What is electric field between the plates​

Answer 1

EXPLANATION.

Let the charge density of plate A can be

calculated as

$\sigma = {17 \times 10 {}^{ - 22} cm {}^{2} }$

Let the charge density of plate B can be

calculate as

$\sigma = { - 17 \times 10 {}^{ - 22}cm {}^{2} }$

To find electric field between the plate,

Electric field between the plate are in the same

direction as total electric field in both the plate

due to surface charge density =

$\frac{ \sigma}{e_{o} }$

Therefore,

electric field on inner side of a plate is =

$\frac{ \sigma}{2 e_{0} }$

For both the plates,

$e \: = \frac{ \sigma}{2 e_{0} } + \frac{ \sigma}{2 e_{0} }$

$e \: = \frac{ \sigma}{ e_{0} }$

put the Value in this equation

$\sigma \: = 17 \times 10 {}^{ - 22}$

$e_{ 0 } \: = permitivity \: of \: \: free \: \: space \: \: = 8.85 \times 10 {}^{ - 12}$

$e \: = \frac{17 \times 10 {}^{ - 22} }{8.85 \times 10 {}^{ - 12} }$

$\boxed{\bold{e \: = 1.92 \: \times 10 {}^{ - 10} \frac{n}{c}}}$