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Question: A conducting sphere of radius \(R\) carrying charge \(Q\) lies inside an uncharged conducting shell ...

A conducting sphere of radius RR carrying charge QQ lies inside an uncharged conducting shell of radius 2R2R. Let us assume that they are joined by a metal wire, then amount of heat that will be developed can be written as,
A.14πε0Q24R B.14πε0Q22R C.14πε0Q2R D.14πε0Q23R \begin{aligned} & A.\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{4R} \\\ & B.\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{2R} \\\ & C.\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{R} \\\ & D.\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{3R} \\\ \end{aligned}

Explanation

Solution

When the conducting spheres of different radius are connected by a metal wire, the whole of the charge would be transferred to the outer conducting shell. Energy inside a capacitor can be found by taking the ratio of the square of the charge inside the shell to twice the capacitance value. These facts will be helpful for you when you are solving this question.

Complete step by step answer:
As we all know, when the conducting spheres of different radius are connected by a metal wire, the whole of the charge would be transferred to the outer conducting shell.
It has been mentioned in the question that the capacitance of the system can be written as,
Ci=4πε0(R){{C}_{i}}=4\pi {{\varepsilon }_{0}}\left( R \right)
Where RR be the radius of the conducting shell.
And the second one will be,
Cf=4πε0(2R){{C}_{f}}=4\pi {{\varepsilon }_{0}}\left( 2R \right)
Therefore, the inertial energy in the capacitor can be written by the equation,
Ei=Q22Ci{{E}_{i}}=\dfrac{{{Q}^{2}}}{2{{C}_{i}}}
And the final energy of the system can be written as,
Ef=Q22Cf{{E}_{f}}=\dfrac{{{Q}^{2}}}{2{{C}_{f}}}
Where QQbe the charge inside the conducting shell.
The amount of heat developed in the shell will be equivalent to the variation in the potential energy of the system. That is we can express this as an equation written as,
Q22CiQ22Cf=14πε0Q24R\dfrac{{{Q}^{2}}}{2{{C}_{i}}}-\dfrac{{{Q}^{2}}}{2{{C}_{f}}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{4R}
Hence the answer has been mentioned in the option as option A.

Note:
Electric potential energy otherwise known as electrostatic potential energy. It is defined as the potential energy which is resulted from coulomb forces which are conservative. This has been related with the configuration of a specific set of point charges within a particular system.