
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
165k+ views
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 –
$q = CV$
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 –
$q = CV$
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
