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.

Step-by-step explanation:

Relation between potential energy and charge is as follows.

U =
`(1)/(4 \pi \epsilon_(o))[(q_(1)q_(2))/(r_(12)) + (q_(2)q_(3))/(r_(23)) + (q_(3)q_(1))/(r_(31))]`

As it is given that
`q_(1) = 1 * 10^(-8) C`,
`q_(2) = 2 * 10^(-8) C`, and
`q_(3) = 3 * 10^(-8) C`.

Distance between the charges = 1 cm =
`1 * 10^(-2) m` (as 1 cm = 0.01 m)

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

U =
`(1)/(4 \pi \epsilon_(o))[(q_(1)q_(2))/(r_(12)) + (q_(2)q_(3))/(r_(23)) + (q_(3)q_(1))/(r_(31))]`

=
`9 * 10^(9) [(1 * 10^(-8) * 2 * 10^(-8))/(10^(-2)) + (2 * 10^(-8) * 3 * 10^(-8))/(10^(-2)) + (3 * 10^(-8) * 1 * 10^(-8))/(10^(-2))]`

=
`9 * 10^(9) [2 + 6 + 1.5]`

=
`85.5 * 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.