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Question: A metallic sphere of radius \(18\,cm \) has been given a charge of \(5 \times {10^{ - 6}}\,C \). The...

A metallic sphere of radius 18cm18\,cm has been given a charge of 5×106C5 \times {10^{ - 6}}\,C . The energy of the charged conductor is:
(A) 0.2J0.2\,J
(B) 0.6J0.6\,J
(C) 1.2J1.2\,J
(D) 2.4J2.4\,J

Explanation

Solution

Hint The energy of the charge can be determined by using the two formulas, the potential difference of the capacitor at a distance formula and then the energy of the charged conductor formula. By using the potential difference of the capacitor formula, the capacitance is determined and then the energy is determined by the energy formula.

Useful formula
The potential difference of the capacitor at a distance is given by,
V=kQrV = \dfrac{{kQ}}{r}
Where, VV is the potential difference, kk is the constant, QQ is the charge of the conductor and rr is the radius.
The energy of the charged conductor is given by,
U=q22CU = \dfrac{{{q^2}}}{{2C}}
Where, UU is the energy of the charged conductor, qq is the charge of the conductor and CC is the capacitance.

Complete step by step answer
Given that,
A metallic sphere of radius is, 18cm=18×102m18\,cm = 18 \times {10^{ - 2}}\,m
The charge of the metallic sphere is, Q=q=5×106CQ = q = 5 \times {10^{ - 6}}\,C
Now,
The potential difference of the capacitor at a distance is given by,
V=kQrV = \dfrac{{kQ}}{r}
By rearranging the terms in the above equation, then the above equation is written as,
QV=rk\dfrac{Q}{V} = \dfrac{r}{k}
Now the term QV\dfrac{Q}{V} is equal to the capacitance of the metallic sphere, then the above equation is written as,
C=rkC = \dfrac{r}{k}
By substituting the radius value and the constant value of the k=9×109k = 9 \times {10^9} in the above equation, then the above equation is written as,
C=18×1029×109C = \dfrac{{18 \times {{10}^{ - 2}}}}{{9 \times {{10}^9}}}
By dividing the terms in the above equation, then the above equation is written as,
C=2×1011C = 2 \times {10^{ - 11}}
Now,
The energy of the charged conductor is given by,
U=q22CU = \dfrac{{{q^2}}}{{2C}}
By substituting the charge and the capacitance in the above equation, then the above equation is written as,
U=(5×106)22×2×1011U = \dfrac{{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}}}{{2 \times 2 \times {{10}^{ - 11}}}}
By squaring the terms in the above equation, then the above equation is written as,
U=2.5×10112×2×1011U = \dfrac{{2.5 \times {{10}^{ - 11}}}}{{2 \times 2 \times {{10}^{ - 11}}}}
By cancelling the same terms in the above equation, then the above equation is written as,
U=2.52×2U = \dfrac{{2.5}}{{2 \times 2}}
By multiplying the terms in the above equation, then the above equation is written as,
U=2.54U = \dfrac{{2.5}}{4}
By dividing the terms in the above equation, then the above equation is written as,
U=0.625JU = 0.625\,J
Thus, the above equation shows the energy of the metallic sphere.

Hence, the option (B) is the correct answer.

Note The capacitance of the conductor is directly proportional to the charge of the conductor and inversely proportional to the potential difference of the conductor. As the charge of the conductor then the capacitance of the conductor also increases.