Solveeit Logo
Login

Question

Question: The capacitance of a variable capacitor can be charged from \(50 \mathrm{pF}\) to \(850 \mathrm{pF}\...

The capacitance of a variable capacitor can be charged from 50pF50 \mathrm{pF} to 850pF850 \mathrm{pF} by turning the dial from 0∘0^{\circ} to 180∘.180^{\circ} . With the dial set at 180∘180^{\circ}, the capacitor is connected to a 400V400 \mathrm{V} After charging the capacitor is disconnected from the battery and the dial is turned to 0∘.0^{\circ} . Now the potential difference across the capacitor is:
(A) 7.6 V
(B) 7.6×103V7.6\times {{10}^{3}}V
(C) 7.6×102V7.6\times {{10}^{2}}V 7.6×102V7.6\times {{10}^{2}}V
(D) None

Explanation

Solution

Hint We know that capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. There are two closely related notions of capacitance: self-capacitance and mutual capacitance. Any object that can be electrically charged exhibits self-capacitance. Capacitance is the ability of a component or circuit to collect and store energy in the form of an electrical charge. Capacitors are energy-storing devices available in many sizes and shapes. It doesn't depend on the EMF of the charging source or on the charges at the plates at some given instant. The charge stored remains the same and thus, one can infer that, the capacitance has increased.

Complete step by step answer From the data given in the question, we know that,
Capacitance when dial is 0∘=500^{\circ}=50 pf =50×10−12F=50 \times 10^{-12} \mathrm{F}
Capacitance when dial is 180∘=850pf=850×10−12F{{180}^{{}^\circ }}=850\text{pf}=850\times {{10}^{-12}}\text{F}
Voltage of the battery = 400V
Energy stored in capacitor,
uC=12CV2=12Q2C{{u}_{C}}=\dfrac{1}{2}C{{V}^{2}}=\dfrac{1}{2}\dfrac{{{Q}^{2}}}{C}
When dial in set at 180∘180^{\circ}
uC=12×850×10−12×(400)2=6.8×10−5J{{u}_{C}}=\dfrac{1}{2}\times 850\times {{10}^{-12}}\times {{(400)}^{2}}=6.8\times {{10}^{-5}}\text{J}
Uc=12Q2C{{U}_{c}}=\dfrac{1}{2}\dfrac{{{Q}^{2}}}{C}
Q=2UCC=2×6.8×10−5×850×10−12Q=\sqrt{2{{U}_{C}}C}=\sqrt{2\times 6.8\times {{10}^{-5}}\times 850\times {{10}^{-12}}}
⇒Q=3.4×10−7C\Rightarrow Q=3.4\times {{10}^{-7}}\text{C}
When dial is set at 0∘0^{\circ}
UC=12Q2C{{U}_{C}}=\dfrac{1}{2}\dfrac{{{Q}^{2}}}{C}
=12×3⋅4×10−7×3⋅4×10−750×10−12=1.44×10−2J=\dfrac{1}{2}\times \dfrac{3\cdot 4\times {{10}^{-7}}\times 3\cdot 4\times {{10}^{-7}}}{50\times {{10}^{-12}}}=1.44\times {{10}^{-2}}\text{J}
1.44×10−2=12CV21.44\times {{10}^{-2}}=\dfrac{1}{2}\text{C}{{\text{V}}^{2}}
⇒V=2×1.44×10−250×10−12=24000V\Rightarrow \text{V}=\sqrt{\dfrac{2\times 1.44\times {{10}^{-2}}}{50\times {{10}^{-12}}}}=24000\text{V}
As no other option matches with the solution.

therefore, the correct answer is Option D.

Note: We can say that if the electric potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1 joule of potential energy when moved between those two locations. Because electric potential difference is expressed in units of volts, it is sometimes referred to as the voltage. Voltmeters are used to measure the potential difference between two points.
There is a misconception about potential and voltage. Many of us think that both are the same. But voltage is not exactly potential; it is the measure of the electric potential difference between two points. When a voltage is connected across a wire, an electric field is produced in the wire. Metal wire is a conductor. Some electrons around the metal atoms are free to move from atom to atom. This causes a difference in energy across the component, which is known as an electrical potential difference.