Solveeit Logo
Login

Question

Question: The potential difference between the plates of a condenser of capacitance \(0.5\mu F\) is \(100\;vol...

The potential difference between the plates of a condenser of capacitance 0.5μF0.5\mu F is 100  volt100\;volt. It is connected to an uncharged condenser of capacity 0.2μF0.2\mu F by a copper wire. The loss of energy in this process will be:
(A) 0  J0\;J
(B) 0.5×10−3J0.5 \times {10^{ - 3}}J
(C) 0.7×10−3J0.7 \times {10^{ - 3}}J
(D) 10−3J{10^{ - 3}}J

Explanation

Solution

Conducting materials have the capacity to store charge. The capacity of a conductor can be defined as the ratio of the charge on the conductor to an increase in potential. The presence of negative charges will reduce the positive potential and capacity of the material will increase. This is the principle of a capacitor.

Formula used:
U=12CVU = \dfrac{1}{2}CV.

Complete step by step solution:
Work has to be done to store charges in a capacitor. This work done will store electrostatic potential in a capacitor.
The capacitance CC is the measure of the ability of a capacitor to store charge.
If VV is the potential difference across the capacitor and CC is the capacitance of the capacitor. Then the energy stored in the capacitor can be written as,
U=12CVU = \dfrac{1}{2}CV
The capacitance is given as
C=0.5μFC = 0.5\mu F
The potential difference is given as
V=100VV = 100V
The initial energy stored can be written as,
U1=12C1V1=120.05×10−6×(100)2=2.5×10−3J{U_1} = \dfrac{1}{2}{C_1}{V_1} = \dfrac{1}{2}0.05 \times {10^{ - 6}} \times {\left( {100} \right)^2} = 2.5 \times {10^{ - 3}}J
Another capacitor is connected to the same potential. Now the potential can be written as,
Vc=C1V1C1+C2{V_c} = \dfrac{{{C_1}{V_1}}}{{{C_1} + {C_2}}}
The capacitance of the second capacitor is given as C2=0.2×10−6F{C_2} = 0.2 \times {10^{ - 6}}F
Substituting in the above equation,
V=(0.5×10−6)×100(0.5×10−6)×(0.2×10−6)=71.43VV = \dfrac{{\left( {0.5 \times {{10}^{ - 6}}} \right) \times 100}}{{\left( {0.5 \times {{10}^{ - 6}}} \right) \times \left( {0.2 \times {{10}^{ - 6}}} \right)}} = 71.43V
The final energy can be written as,
Uf=12(C1+C2)VC2{U_f} = \dfrac{1}{2}\left( {{C_1} + {C_2}} \right)V_C^2
Substituting the values we get,
Uf=12(0.5+0.2)×10−6×(71.43)2=1.8×10−3J{U_f} = \dfrac{1}{2}\left( {0.5 + 0.2} \right) \times {10^{ - 6}} \times {\left( {71.43} \right)^2} = 1.8 \times {10^{ - 3}}J
The energy loss through the wire can be written as the difference between the initial and final energies.
U=Uf−UiU = {U_f} - {U_i}
Substituting, we get
U=(2.5−1.8)×10−3=0.7×10−3JU = \left( {2.5 - 1.8} \right) \times {10^{ - 3}} = 0.7 \times {10^{ - 3}}J
The answer is: Option (C): 0.7×10−3J0.7 \times {10^{ - 3}}J.

Note:
The capacitance is defined as the amount of charge required to increase the potential by one volt. Capacitors are used to store charges, to generate electromagnetic oscillations, to tune radio circuits, to reduce voltage fluctuations in power supply and to eliminate sparking in the ignition system of automobile engines.