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How much kinetic energy will be gained by an $α$ – particle in going from a point at 70 V to another point at 50 V ?

$40 e V$

$40 k e V$

$40 M e V$

$0 e V$

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Explanation

$K E = q (V_{1} - V_{2}) = 2 \times 1 .6 \times 10^{- 19} \times (70 - 50) = 40 e V$

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If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be -

$\frac{1}{3} V$

$\frac{2}{3} V$

$\frac{3}{2} V$

3 V

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Explanation

Potential inside the sphere will be same as that on its surface i.e. $V = V_{s u r f a c e} = \frac{K Q}{10} u n i t s$ , $V_{o u t} = \frac{K Q}{15} u n i t s$

∴ $\frac{V_{o u t}}{V} = \frac{2}{3}$ ⇒ $V_{o u t} = \frac{2}{3} V$

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What is the potential energy of the equal positive point charges of 1 μC each held 1 m apart in air ?

$9 \times 10^{- 3} J$

$9 \times 10^{- 3} e V$

$2 e V / m$

Zero

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Explanation

By using $U = 9 \times 10^{9} \frac{Q_{1} Q_{2}}{r}$

⇒ $U = 9 \times 10^{9} \times \frac{10^{- 6} \times 10^{- 6}}{1} = 9 \times 10^{- 3} J$

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An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m³, the radius of the drop will be -

$2 .0 \times 10^{- 6} m$

$1 .7 \times 10^{- 6} m$

$1 .4 \times 10^{- 6} m$

$1 .1 \times 10^{- 6} m$

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Explanation

In equilibrium QE = mg ⇒ $Q . \frac{V}{d} = m g = (\frac{4}{3} π r^{3} ρ) g$

⇒ $2 \times 1 .6 \times 10^{- 19} \times \frac{12000}{2 \times 10^{- 2}} = \frac{4}{3} π r^{3} \times 900 \times 10$

⇒ r = 1.7 × 10^–6 m

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The ratio of momenta of an electron and an $α$ -particle which are accelerated from rest by a potential difference of 100 volt is

$\sqrt{\frac{2 m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{2 m_{α}}}$

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Explanation

Momentum $p = \sqrt{2 m K};$ where K = kinetic energy = Q.V

⇒ $p = \sqrt{2 m Q V}$ ⇒ $p \propto \sqrt{m Q}$

$\Rightarrow \frac{p_{e}}{p_{α}} = \sqrt{\frac{m_{e} Q_{e}}{m_{α} Q_{α}}} = \sqrt{\frac{m_{e}}{2 m_{α}}}$

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When a proton is accelerated through 1V, then its kinetic energy will be -

1840 eV

13.6 eV

1 eV

0.54 eV

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Explanation

$Δ K E = q V = e V = e \times 1 = 1 e V$

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Ten electrons are equally spaced and fixed around a circle of radius R. Relative to V = 0 at infinity, the electrostatic potential V and the electric field E at the centre C are

$V \neq 0$ and $\vec{E} \neq 0$

$V \neq 0$ and $\vec{E} = 0$

V = 0 and $\vec{E} = 0$

V = 0 and $\vec{E} \neq 0$

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Explanation

At centre E = 0, V ≠ 0 because the electric field is a vector quantity but the electric potential is not a vector quantity.

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The displacement of a charge Q in the electric field $E = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\hat{r} = a \hat{i} + b \hat{j}$ . The work done is

$Q (a e_{1} + b e_{2})$

$Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}$

$Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}$

$Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)$

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Explanation

By using $W = Q (\vec{E} . Δ \vec{r})$

⇒ $W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

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A cube of a metal is given a positive charge Q. For the above system, which of the following statements is true

Electric potential at the surface of the cube is zero

Electric potential within the cube is zero

Electric field is normal to the surface of the cube

Electric field varies within the cube

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Explanation

Electric lines of force are always normal to metallic body.

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Electric potential at any point is $V = - 5 x + 3 y + \sqrt{15} z$ , then the magnitude of the electric field is

$3 \sqrt{2}$

$4 \sqrt{2}$

$5 \sqrt{2}$

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Explanation

$E_{x} = - \frac{d V}{d x} = - (- 5) = 5;$ $E_{y} = - \frac{d V}{d y} = - 3$

and $E_{z} = - \frac{d V}{d z} = - \sqrt{15}$

$E_{n e t} = \sqrt{E_{x}^{2} + E_{y}^{2} + E_{z}^{2}} = \sqrt{{(5)}^{2} + {(- 3)}^{2} + {(- \sqrt{15})}^{2}} = 7$

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