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Question: The capacitance of a variable ratio capacitor can be charged from \(50pF\) to \(950pF\) by tuning th...

The capacitance of a variable ratio capacitor can be charged from 50pF50pF to 950pF950pF by tuning the dial from 0 to 180∘{180^ \circ }. When the dial is set at 180∘{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?

Explanation

Solution

Hint: - Convert the unit of capacitance to farad when the dial is set at 0 and 180∘{180^ \circ } from 50pF50pF to 950pF950pF. Find the energy stored in the capacitor when the dial is set at 0 and 180∘{180^ \circ }. Find the difference between the energies stored in the capacitor when the dial is set at 0 and 180∘{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=50×10−12F = 50pF = 50 \times {10^{ - 12}}F
Capacitance when the dial is at 180∘=950pF=950×10−12F{180^ \circ } = 950pF = 950 \times {10^{ - 12}}F
Voltage of the battery =400V = 400V
Now, we know that, energy stored in the capacitor is given by –
U=12CV2U = \dfrac{1}{2}C{V^2}
where, CC is the capacitance and VV is the voltage
Now, when the dial is set to 180∘{180^ \circ } then energy stored –
⇒U=12×950×10−12×(400)2 ∴U=7.6×10−5J  \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 –
⇒U=12CV2 ⇒U=Q22C  \Rightarrow U = \dfrac{1}{2}C{V^2} \\\ \Rightarrow U = \dfrac{{{Q^2}}}{{2C}} \\\
∴Q=2UC\therefore Q = \sqrt {2UC}
Putting the value in the above formula, we get –
⇒Q=2×7.6×10−5×950×10−12 ∴Q=3.8×10−3C  \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=Q22C ⇒U=12×(3.8×10−3)250×10−12 ∴U=1.444×10−3J  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 –
⇒1.444×10−3=12CV2 ⇒V=2×1.444×10−350×10−12 ∴V=7600V  \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∘{180^ \circ } -
ΔU=1.444×10−3−7.6×10−5 ⇒ΔU=1.368×10−3J  \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 CC then the relation between charge, capacitance and potential difference is given by –
q=CVq = CV