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?

Question

asked May 15, 2020 34.3k views

1 Answer

Galeksic · Answer 1 · 2020-05-20T16:55:12+0000

Answer:

The charge is
$2.75*10^(-13)\ C$

Step-by-step explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length
$L=5.0*5.0*10^(-4)\ m$

We need to calculate the linear charge density

Using formula of linear charge density

$E=(2k\lambda)/(r)$

$\lambda=(Er)/(2k)$

Put the value into the formula

$\lambda=(800*2.5*10^(-3))/(2*9*10^(9))$

$\lambda=1.1*10^(-10)\ C/m$

We need to calculate the charge

Using formula of charge

$Q=\lambda*L$

Put the value into the formula

Q=1.1*10^(-10)*(5.0*5.0*10^(-4))

$Q=2.75*10^(-13)\ C$

Hence, The charge is
$2.75*10^(-13)\ C$