Question

The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of the charged sheet?

Answer 1

Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length $L=5.0\times5.0\times10^{-4}\ m$

We need to calculate the linear charge density

Using formula of linear charge density

$E=\dfrac{2k\lambda}{r}$

$\lambda=\dfrac{Er}{2k}$

Put the value into the formula

$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

$\lambda=1.1\times10^{-10}\ C/m$

We need to calculate the charge

Using formula of charge

$Q=\lambda\timesL$

Put the value into the formula

$Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})$

$Q=2.75\times10^{-13}\ C$

Hence, The charge is $2.75\times10^{-13}\ C$