Question

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?

Answer 1

Answer:

(a) The charge on capacitor in first case is $8.5 \times 10^{-11}$ C

(b) In second case is $4.25 \times 10^{-11}$ C

(c) Potential across plate is 24 V

(d) Work required to pull the plates is $5.1 \times 10^{-10}$ J

Explanation:

Given:

Area of plate $A = 4 \times 10^{-4}$ $m^{2}$

Voltage $V = 12$ V

Separation between two plate $d = 0.50 \times 10^{-3}$ m

(a)

The charge on capacitor is given by,

   $C = \frac{Q}{V}$

But capacitance of parallel plate capacitor is given by,

   $C = \frac{\epsilon_{o} A }{d}$

⇒ $Q = \frac{\epsilon _{o}A V }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

   $Q = \frac{8.85\times 10^{-12} \times 4 \times 10^{-4} \times 12}{0.50 \times 10^{-3} }$

   $Q = 8.49 \times 10^{-11}$ C

(b)

The charge on the capacitor when plate separation is $1 \times 10^{-3}$ m is,

Here $d = 1 \times 10^{-3}$ m

     $Q = \frac{\epsilon _{o}A V }{d}$

     $Q = \frac{8.85\times 10^{-12} \times 4 \times 10^{-4} \times 12}{1 \times 10^{-3} }$

     $Q = 4.25 \times 10^{-11}$ C

(c)

The potential difference across plate is,

     $V = \frac{Q}{C}$

But $C = \frac{\epsilon_{o} A }{d}$

  $C = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4} }{1 \times 10^{-3} }$

  $C = 3.45 \times 10^{-12}$ F

Put the value of capacitance and find potential difference,

  $V = \frac{8.5 \times 10^{-11} }{3.45 \times 10^{-12} }$

  $V = 24$ V

(d)

Work required is given by,

   $W= U_{f} - U_{i}$

   $W = \frac{Q (V_{f} -V_{i} )}{2}$

Where $V_{f} = 24$ V and $V_{i} = 12$

   $W = \frac{8.5 \times 10^{-11} \times 12}{2}$

  $W = 5.1 \times 10^{-10}$ J

Therefore, the charge on capacitor in first case is $8.5 \times 10^{-11}$ C and in second case is $4.25 \times 10^{-11}$ C and potential across plate is 24 V and work required to pull the plates is $5.1 \times 10^{-10}$ J