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?

Charge,

Explanation:

It is given that,

Value of electric field,

Area of parallel plates,

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

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

.......(1)

Since,

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

And .....(3)

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

So, the charge on the each plate is . Hence, this is the required solution.

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