Question

The capacitance of a variable ratio capacitor can be charged from $50pF$ to $950pF$ by tuning the dial from 0 to ${180^ \circ }$. When the dial is set at ${180^ \circ }$ then the capacitor is connected to a 400V battery. When it is charged the capacitor is disconnected from the battery and the dial is turned to 0. (a) When the dial is set at 0 what is the potential difference across the capacitor? (b) How much work is required to turn the dial given that the friction is neglected?

Answer 1

Hint: - Convert the unit of capacitance to farad when the dial is set at 0 and ${180^ \circ }$ from $50pF$ to $950pF$. Find the energy stored in the capacitor when the dial is set at 0 and ${180^ \circ }$. Find the difference between the energies stored in the capacitor when the dial is set at 0 and ${180^ \circ }$ to find the net work done.

Complete Step by Step Solution: -
According to the question, it is given that –
Capacitance when the dial is at 0 $ = 50pF = 50 \times {10^{ - 12}}F$
Capacitance when the dial is at ${180^ \circ } = 950pF = 950 \times {10^{ - 12}}F$
Voltage of the battery $ = 400V$
Now, we know that, energy stored in the capacitor is given by –
$U = \dfrac{1}{2}C{V^2}$
where, $C$ is the capacitance and $V$ is the voltage
Now, when the dial is set to ${180^ \circ }$ then energy stored –
$
   \Rightarrow U = \dfrac{1}{2} \times 950 \times {10^{ - 12}} \times {\left( {400} \right)^2} \\
  \therefore U = 7.6 \times {10^{ - 5}}J \\
 $
Using the formula of energy stored in the capacitor to find the charge across the capacitor –
$
   \Rightarrow U = \dfrac{1}{2}C{V^2} \\
   \Rightarrow U = \dfrac{{{Q^2}}}{{2C}} \\
 $
$\therefore Q = \sqrt {2UC} $
Putting the value in the above formula, we get –
$
   \Rightarrow Q = \sqrt {2 \times 7.6 \times {{10}^{ - 5}} \times 950 \times {{10}^{ - 12}}} \\
  \therefore Q = 3.8 \times {10^{ - 3}}C \\
 $
Now, when the dial is set at 0 then, using the formula of energy stored and putting the values in it, we get –
$
  U = \dfrac{{{Q^2}}}{{2C}} \\
   \Rightarrow U = \dfrac{1}{2} \times \dfrac{{{{\left( {3.8 \times {{10}^{ - 3}}} \right)}^2}}}{{50 \times {{10}^{ - 12}}}} \\
  \therefore U = 1.444 \times {10^{ - 3}}J \\
 $
To find the potential difference across the capacitor use the formula of energy stored we get –
$
   \Rightarrow 1.444 \times {10^{ - 3}} = \dfrac{1}{2}C{V^2} \\
   \Rightarrow V = \sqrt {\dfrac{{2 \times 1.444 \times {{10}^{ - 3}}}}{{50 \times {{10}^{ - 12}}}}} \\
  \therefore V = 7600V \\
 $
Hence, the potential difference across the capacitor is 7600V.
Work done can be calculated by finding the difference between the energy stored in the capacitor when the dial is set at 0 and ${180^ \circ }$ -
$
  \Delta U = 1.444 \times {10^{ - 3}} - 7.6 \times {10^{ - 5}} \\
   \Rightarrow \Delta U = 1.368 \times {10^{ - 3}}J \\
 $

Note: - The energy stored in the capacitor is the electric potential energy and is related to the voltage and charge on the capacitor. If the capacitance of the conductor is $C$ then the relation between charge, capacitance and potential difference is given by –
$q = CV$