The value of electric potential at a distance of 9 cm from the point charge 4 × 10⁻⁷ C is [Given (1)/(4πε₀) = 9 × 10⁹ N m²C⁻² ]: (a) 4 × 10² V (b) 44.4 V (c) 4.4 × 10⁵ V (d) 4 × 10⁴ V

Answer: (d). V = kq/r = 9 × 10⁹ × 4 × 10⁻⁷0.09 V = (3600)/(0.09) = 4 × 10⁴ V So the answer is (d) 4 × 10⁴ V.

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Electric Potential & Equipotentials

The value of electric potential at a distance of $9\text{ cm}$ from the point charge $4 \times 10^{-7}\text{ C}$ is [Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2\text{C}^{-2}$ ]:

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Answer: (d)

$V = \frac{kq}{r} = \frac{9 \times 10^{9} \times 4 \times 10^{-7}}{0.09}$
$V = \frac{3600}{0.09} = 4 \times 10^{4}\text{ V}$
So the answer is (d) $4 \times 10^{4}\text{ V}$ .

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