Question

The potential energy of system of two equal negative point charges of 2μC each held 1 m apart in air is (k=\[9 \times {10^9}\]SI unit)
(A) 36\[J\]
(B) \[3.6 \times {10^{ - 3}}J\]
(C) \[3.6J\]
(D) \[3.6 \times {10^{ - 2}}J\]

Answer 1

Hint: The potential energy is defined as the amount of work done in moving a unit positive charge from infinity to that point without undergoing any acceleration. It is a scalar quantity and has no direction. For this sum, the potential energy is found out by work done by the force between the charges based on Coulomb's electrostatic law for the two charges that are separated by a distance r.
Formula used: The potential energy of a system is given by,
\[U = k\dfrac{{{q_1}{q_2}}}{r}\]
where,
U is the potential energy of the system
 k is a proportionality constant equal to\[9 \times {10^9}\]
 \[{q_1}\]and \[{q_2}\] are the two point charges of the system
  r is the distance of separation between the two point charges.

Complete step by step solution: Given that the system has two equal negative point charges. So,
 \[{q_1}\]=\[{q_2}\]=-2μC=\[ - 2 \times {10^{ - 6}}C\]and r=1m
Substituting the given values in the potential energy expression, we get,
\[U = 9 \times {10^9} \times \dfrac{{ - 2 \times {{10}^{ - 6}} \times - 2 \times {{10}^{ - 6}}}}{1}\]
\[U = 9 \times {10^9} \times 4 \times {10^{ - 12}}\]
\[U = 36 \times {10^{ - 3}}\]
\[U = 3.6 \times {10^{ - 2}}J\]
Thus, the potential energy of a system of two equal negative point charges of 2μC each held 1 m apart in air is\[3.6 \times {10^{ - 2}}J\].

The correct option is D.

Note: The potential energy of the system increases if two like charges are brought towards each other while the potential energy of the system decreases when two unlike charges are brought towards each other. Also, only for conservative force, the potential energy can be defined.