Important NEET PYQs – Electrostatic Potential and Capacitance

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Topic Question Set

Q 1 :
The value of electric potential at a distance of 9 cm from the point charge $4 \times 10^{- 7} C$ is

$(Given that \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2})$ [2024]

$4 \times 10^{2}$ $V$
$44.4$ $V$
$4.4 \times 10^{5}$ $V$
$4 \times 10^{4}$ $V$

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(4)

$r = 9$ $cm = 9 \times 10^{- 2}$ $m$

$q = 4 \times 10^{- 7}$ $C$

As, $V = \frac{1}{4 π ε_{0}} \frac{q}{r} = \frac{9 \times 10^{9} \times 4 \times 10^{- 7}}{9 \times 10^{- 2}} = 4 \times 10^{4}$ $V$

Q 2 :
If a conducting sphere of radius $R$ is charged. Then the electric field at a distance $r (r > R)$ from the centre of the sphere would be, (V = potential on the surface of the sphere) [2023]

$\frac{r V}{R^{2}}$
$\frac{R^{2} V}{r^{3}}$
$\frac{R V}{r^{2}}$
$\frac{V}{r}$

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(3)

Consider a sphere of radius $' R'$ and charge $' Q'$ .

Given, $r > R$

Potential, $V = \frac{K Q}{R}$ ...(i)

Electric field, $E = \frac{K Q}{r^{2}}$ ...(ii)

Using value of $Q = \frac{V R}{K}$ in equation (ii),

$E = \frac{K}{r^{2}} \times \frac{V R}{K} = \frac{V R}{r^{2}}$

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