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GUJCET · Physics · Electrostatic Potential and Capacitance
Q amount of electric charge is present on the surface of a sphere having radius R. Then electrical potential energy of this system is _______.
- A \(\frac{k Q ^2}{ R }\)
- B \(\frac{k Q^2}{R^2}\)
- C \(\frac{1}{2} \frac{ kQ ^2}{ R }\)
- D \(\frac{1}{2} \frac{k Q^2}{R^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} \frac{ kQ ^2}{ R }\)
Step-by-step Solution
Detailed explanation
(C)
Initially potential when there is no charge on the sphere,
\(V_1=0\)
potential when there is Q electric charge on the sphere,
\(V_2=\frac{k Q}{R}\)
average potential \(V =\frac{ V _1+ V _2}{2}\)
\(\begin{array}{l}\therefore V =\frac{ O +\frac{k Q }{ R }}{2} \\ \therefore V=\frac{k Q }{2 R }\end{array}\)
potential energy of the system \(= QV\)
\(\begin{array}{l} U = Q \left(\frac{k Q }{2 R }\right) \\ U =\frac{1}{2} \frac{k Q ^2}{ R }\end{array}\)
Initially potential when there is no charge on the sphere,
\(V_1=0\)
potential when there is Q electric charge on the sphere,
\(V_2=\frac{k Q}{R}\)
average potential \(V =\frac{ V _1+ V _2}{2}\)
\(\begin{array}{l}\therefore V =\frac{ O +\frac{k Q }{ R }}{2} \\ \therefore V=\frac{k Q }{2 R }\end{array}\)
potential energy of the system \(= QV\)
\(\begin{array}{l} U = Q \left(\frac{k Q }{2 R }\right) \\ U =\frac{1}{2} \frac{k Q ^2}{ R }\end{array}\)
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