Q139 / 180 · KCET 2015 · Physics
Three point charges \( 3 n C .6 n C \) and \( 9 n C \) are placed at the comers of an equilateral triangle of
side \( 0.1 \mathrm{~m} \). The potential energy of the system is
side \( 0.1 \mathrm{~m} \). The potential energy of the system is
- A\( 8910 \mathrm{~J} \)
- B\( 9910 \mathrm{~J} \)
- CO \( 99100 \mathrm{~J} \)
- Correct answerDNone of above
Solution
Potential energy is given as \( U=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \)
The three charges are placed at the corners of an equilateral triangle as shown in the following figure. The potential
energies of the three charges are given as
\[ \begin{aligned} U_{1} &=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{3 n C \times 6 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{18 \times 10^{-18}}{0.1} \\ U_{2} &=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{6 n C \times 9 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{54 \times 10^{-18}}{0.1} \\ U_{3} &=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{9 n C \times 3 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{27 \times 10^{-18}}{0.1} \end{aligned} \]
Therefore, potential energy of the system is given as
\[ \begin{array}{l} U=U_{1}+U_{2}+U_{3}=\frac{1}{4 \Pi \varepsilon_{0}}\left[\frac{18+54+27}{0.1}\right] \times 10^{-18} \\ =\frac{9 \times 10^{9} \times 99 \times 10^{-18}}{0.1}=8910 \times 10^{-9} \mathrm{~J} \end{array} \]
The three charges are placed at the corners of an equilateral triangle as shown in the following figure. The potential
energies of the three charges are given as
\[ \begin{aligned} U_{1} &=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{3 n C \times 6 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{18 \times 10^{-18}}{0.1} \\ U_{2} &=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{6 n C \times 9 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{54 \times 10^{-18}}{0.1} \\ U_{3} &=\frac{1}{4 \Pi \varepsilon_{0}} \times \frac{9 n C \times 3 n C}{0.1}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{27 \times 10^{-18}}{0.1} \end{aligned} \]
Therefore, potential energy of the system is given as
\[ \begin{array}{l} U=U_{1}+U_{2}+U_{3}=\frac{1}{4 \Pi \varepsilon_{0}}\left[\frac{18+54+27}{0.1}\right] \times 10^{-18} \\ =\frac{9 \times 10^{9} \times 99 \times 10^{-18}}{0.1}=8910 \times 10^{-9} \mathrm{~J} \end{array} \]
Hence, the correct answer is option: 'D'.