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

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`