What is the absolute potential at point ‘P’ which is 2m from a point charge Q = + 5μC and the work required to move a +8 nC charge from infinity to P?

Assume ϵ_{r} = 1

This question was previously asked in

UGC NET Paper 2 (Electronic Science) December 2019 Official Paper

Option 4 : 22.5 KV, 180 μJ

Official Paper 1: Held on 24 Sep 2020 Shift 1

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50 Questions
100 Marks
60 Mins

__Concept__:

Absolute potential (V) denotes the amount of work done to bring a unit positive charge from infinity to a point.

Mathematically, V is given by:

\(V = \frac{{kq}}{r}\)

\(k = \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\;N{m^2}/{C^2}\)

Q = Charge

r = Distance from the charge

On multiplying the potential with the amount of charge, we get the potential energy which is the amount of work done in bringing a charge ‘Q’ from infinity to a point, i.e.

\(W = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}r}}\)

__Calculation__:

Given q = 5 nC and r = 2 m

\(V = 9 \times {10^9} \times \frac{{5 \times {{10}^{ - 6}}}}{2}\)

V = 22.5 kV

Now, with q_{1} = +5μc and q_{2} = +8nc, the Work done to move a +8 nC charge from infinity to a point (P) is given by the product of the charge and the potential, i.e.

\(W = \frac{{9 \times {{10}^9} \times 5 \times {{10}^{ - 6}} \times 8 \times {{10}^{ - 9}}}}{2}J\)

U = 180 μJ