Question

Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3 = 3 cm. The potential energy of this arrangement, relative to the potential energy for infinite separation, is about:
1. 0J2. 0.16J3. 0.079 J4. 0.00085 J5. 0.0017J

Answer 1

Answer: Option (4) is the correct answer.

Explanation:

Relation between potential energy and charge is as follows.

                 U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

As it is given that $q_{1} = 1 \times 10^{-8} C$, $q_{2} = 2 \times 10^{-8} C$, and $q_{3} = 3 \times 10^{-8} C$.

        Distance between the charges = 1 cm = $1 \times 10^{-2} m$  (as 1 cm = 0.01 m)

Hence, putting these given values into the above formula as follows.

                 U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

            = $9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]$    

            = $9 \times 10^{9} [2 + 6 + 1.5]$

            = $85.5 \times 10^{-5}$ J

            = 0.00085 J

Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.