Two charges q1 = 4 μC and q2 = 16 μC are placed 9 cm apart. At a point P between them, the electric field is zero. If a charge of 5 μC is placed at P, the potential energy of the system is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 24.4 J
The field between two like charges is zero at the point where their fields are equal and opposite. The potential energy of a system of point charges is the sum over every pair of charges.
Let P be at a distance x from \( q_1 \), so it is \( 9 - x \) cm from \( q_2 \).
\( \displaystyle \frac{kq_1}{x^{2}} = \frac{kq_2}{(9 - x)^{2}} \)
\( \displaystyle \frac{9 - x}{x} = \sqrt{\frac{q_2}{q_1}} = \sqrt{\frac{16}{4}} = 2 \)
\( x = 3\,\mathrm{cm} = 0.03\,\mathrm{m} \)
\( 9 - x = 6\,\mathrm{cm} = 0.06\,\mathrm{m} \)
With q = 5 μC at P and all three charges in μC (so each product of two charges carries \( 10^{-12}\,\mathrm{C^{2}} \)):
\( \displaystyle U = k\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q}{r_{1P}} + \frac{q_2q}{r_{2P}}\right) \)
\( \displaystyle U = (9\times10^{9}) \times 10^{-12}\left(\frac{4 \times 16}{0.09} + \frac{4 \times 5}{0.03} + \frac{16 \times 5}{0.06}\right)\,\mathrm{J} \)
\( U = 9\times10^{-3} \times (711.1 + 666.7 + 1333.3)\,\mathrm{J} \)
\( U = 9\times10^{-3} \times 2711.1\,\mathrm{J} \approx 24.4\,\mathrm{J} \)