A metal plate P is placed symmetrically between the plates A and B of a capacitor of capacitance `C_(0)` as shown in the figure. Plate A is given a charge Q and plate B is given a charge `3Q`. If `V_(B)-V_(A)=a((Q)/(C_(0)))`, a is equal to _____.

Answer

1

Explanation

Let the separation between the plates A and B be 2d.
Then, `C_(0)=(epsi_(0)A)/(2d)`
The charge distribution on the six surfaces of the plates will be as shown.

So, `V_(B)-V_(P)=((Q)/(epsi_(0)A))d=(Q)/(2C_(0))`
And, `V_(P)-V_(A)=(Q)/(epsi_(0)A)d=(Q)/(2C_(0))`.
Adding the above two equations, `V_(B)-V_(A)=(Q)/(C_(0))`.