A 100pF capacitor is charged to a potential difference of 20V. It is connected to an uncharged capacitor of capacitance 20pF. What will be the new potential difference across the 100pF capacitor?

Hint: When the first capacitor is connected across a potential difference, it gets charged up. Now when it is connected in parallel with another capacitor, the charge will be distributed between the two. Since there is no external supply of charge, the total charge before and after addition of the new capacitor will be the same.


Formula Used:
1.Total charge stored by capacitor:$Q = CV$ ……(1)
where,
C is the capacitance of the capacitor
V is the voltage across the capacitor
2. $1F = {10^{12}}pF$ (conversion of 1 Farad(F) into Pico farad(pF))


Complete step by step answer:
Given:
1. Potential difference across capacitor: V=20 V
2. Capacitance of first capacitor: ${C_1} = 100pF$
3. Capacitance of second capacitor: ${C_2} = 20pF$

To find: The new potential difference across the 100pF capacitor.

Step 1:
Convert capacitance of both the capacitors to F:
$
  {C_1} = 100 \times {10^{ - 12}}F \\
   \Rightarrow {C_1} = {10^{ - 10}}F \\
 $
$
  {C_2} = 20 \times {10^{ - 12}}F \\
   \Rightarrow {C_2} = 0.2 \times {10^{ - 10}}F \\
 $
Find the total charge stored in capacitor initially using eq (1):
\[
  {Q_{initial}} = {C_1}V \\
   \Rightarrow {Q_{initial}} = {10^{ - 10}} \times 20C \\
   \Rightarrow {Q_{initial}} = 2 \times {10^{ - 9}}C = 2nC \\
 \]

Step 2:
As no new charge is added to the circuit, the total charge before and after addition of the second capacitor will be the same.
${Q_{initial}} = {Q_{final}}$
The final charge is the addition of the charge stored in both the capacitors:
$
  {Q_{final}} = {Q_1} + {Q_2} \\
   \Rightarrow {Q_{initial}} = {Q_1} + {Q_2} \\
 $

Step 3:
As the two capacitors are connected in parallel, the voltage across the two will be the same. Substitute the expression for Q using eq (1):
$
  {Q_{initial}} = {C_1}V + {C_2}V \\
   \Rightarrow {Q_{initial}} = (1 + 0.2) \times {10^{ - 10}}V \\
 $
Substitute the value of initial Q and rearrange to find V:
$
  2 \times {10^{ - 9}} = (1.2) \times {10^{ - 10}}V \\
   \Rightarrow V = \dfrac{{2 \times {{10}^{ - 9}}}}{{(1.2) \times {{10}^{ - 10}}}} \\
   \Rightarrow V = 0.167V \\
 $
As the voltage across both the capacitors is the same the voltage across the 100pF capacitor will be 0.167V.

Final Answer:
The voltage across the 100pF capacitor will be 0.167V.


Note: The capacitance tells us how much charge the device stores for a given voltage.
For two capacitors in parallel, the voltage across them will be the same but the charge stored will be divided in order to conserve the total charge. The one with the higher capacitance will store more charge and energy than the one with lower capacitance for the same voltage.