The electric field and electric potential at any point due to a point charge kept in air are $20\ \text{N C}^{-1}$ and $10\ \text{J C}^{-1}$ respectively. Compute the magnitude of this charge.

$q = 0.55\ \text{nC}\ (5.5 \times 10^{-10}\ \text{C})$.

CBSE Class 12 · 2006 · ElectrostaticsShort Answer

The electric field and electric potential at any point due to a point charge kept in air are

20 N C^{- 1}

and

10 J C^{- 1}

respectively. Compute the magnitude of this charge.

Answer: $q = 0.55 nC (5.5 \times 1 0^{- 10} C)$ .

Solution

The field and potential at the point are given:

E = \frac{k q}{r ^{2}} = 20, V = \frac{k q}{r} = 10

Dividing the two:

\frac{V}{E} = r = \frac{10}{20} = 0.5 m

Then:

q = \frac{V r}{k} = \frac{10 \times 0.5}{9 \times 1 0 ^{9}}

q = 5.5 \times 1 0^{- 10} C = 0.55 nC

CBSE Class 12 Board · 2006 · Subjective

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