Question

An electric field of 8.5X10^5 V/m is desired between two parallel plates each of areas 2500 cm^2 and separated by 0.10 mm of air. What charge must be on each plate?

Answer 1

The charge must be on each plate is
`1.881*10^(-6)\ C`.

Step-by-step explanation:

Given that,

Electric field
`E= 8.5*10^(5)\ V/m`

Area = 2500 cm²

Distance = 0.10 mm

We need to calculate the potential difference

Using formula of potential difference

`\Delta V=E* d`

`\Delta V=8.5*10^(5)*0.10*10^(-3)`

`\Delta V=85\ V`

We need to calculate the capacitor

Using formula of capacitor

`C=(\epsilon_(0)* A)/(d)`

Put the value into the formula

`C=(8.85*10^(-12)*2500*10^(-4))/(0.10*10^(-3))`

`C=2.2125*10^(-8)\ F`

We need to calculate the charge

Using formula of charge

`Q=C*\Delta V`

`Q=2.2125*10^(-8)*85`

`Q=1.881*10^(-6)\ C`

Hence, The charge must be on each plate is
`1.881*10^(-6)\ C`.