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Electrostatics Part 3

Section: Physics

Question : 12 of 80

Marks: +1, -0

\_\_\_\_

\mu\,\mathrm{J}

\; \frac{1}{4\pi E_0} = 9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2 / \mathrm{C}^2

Solution:

Concept:

Work done in bringing a charge from infinity to a point equals the product of the charge and the electric potential at that point.

Explanation:

Let

q_1 = 1\ \mathrm{nC}

q_2 = 2\ \mathrm{nC}

q_3 = 3\ \mathrm{nC}

, and side

r = 3\ \mathrm{cm} = 0.03\ \mathrm{m}

Potential at the third corner due to

q_1

and

q_2

V_f = \dfrac{k q_1}{r} + \dfrac{k q_2}{r} = \dfrac{k (q_1+q_2)}{r}

Initial potential at infinity is

V_i = 0

Work done:

W = q_3 \times (V_f - V_i) = q_3 \times \dfrac{k (q_1+q_2)}{r}

Substitute values:

k = 9 \times 10^9\ \mathrm{N\cdot m^2/C^2}

q_3 = 3\times10^{-9}\ \mathrm{C}

q_1+q_2 = 3\times10^{-9}\ \mathrm{C}

r = 0.03\ \mathrm{m}

W = \dfrac{(9\times10^9) \times (3\times10^{-9}) \times (3\times10^{-9})}{0.03}

Simplify:

W = \dfrac{27 \times 3 \times 10^{-9}}{0.03} = \dfrac{81\times10^{-9}}{0.03} = 27\times10^{-7}\ \mathrm{J}

Convert:

1\ \mathrm{J} = 10^6\ \mu\mathrm{J}

, so

W = 27\times10^{-7} \times 10^6\ \mu\mathrm{J} = 27\times10^{-1}\ \mu\mathrm{J} = 2.7\ \mu\mathrm{J}

Answer:

2.7\ \mu\mathrm{J}

, which corresponds to option (C).

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