A parallel plate capacitor with plate area 4.0 cm2 and air gap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected.
(a) What is the charge on the capacitor?
(b) The plates are now pulled to a separation of 1.00 mm. What is the charge on the capacitor now?

(c) What is the potential difference across the plates now?

(d) How much work was required to pull the plates to their new separation?

(a) The charge on capacitor in first case is C

(b) In second case is C

(c) Potential across plate is 24 V

(d) Work required to pull the plates is J

Explanation:

Given:

Area of plate

Voltage V

Separation between two plate m

(a)

The charge on capacitor is given by,

But capacitance of parallel plate capacitor is given by,

⇒

Where

    C

(b)

The charge on the capacitor when plate separation is m is,

Here m

      C

(c)

The potential difference across plate is,

But

  F

Put the value of capacitance and find potential difference,

  V

(d)

Work required is given by,

Where V and

  J

Therefore, the charge on capacitor in first case is C and in second case is C and potential across plate is 24 V and work required to pull the plates is J

f=ma newton

f=100x2.5=250n

god i hope this !