Question

An electric field of 7.50×105 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?

Answer 1

Answer:

Charge, $q=2.98\times 10^{-8}\ C$

Explanation:

It is given that,

Value of electric field, $E=7.5\times 10^5\ V/m$

Area of parallel plates, $A=45\ cm^2=0.0045\ m^2$

Distance between two parallel plates, d = 2.45 mm = 0.00245 m

For a parallel plate capacitor, the capacitance is given by :

$C=\dfrac{\epsilon_oA}{d}$.......(1)

Since, $E=\dfrac{V}{d}$

V = E . d ............(2)

And $C=\dfrac{q}{V}$.....(3)

From equation (1), (2) and (3) we get :

$\dfrac{q}{V}=\dfrac{\epsilon_oA}{d}$

$q=\epsilon_o EA$

$q=8.85\times 10^{-12}\ F/m\times 7.5\times 10^5\ V/m\times 0.0045\ m^2$

$q=2.98\times 10^{-8}\ C$

So, the charge on the each plate is $2.98\times 10^{-8}\ C$. Hence, this is the required solution.