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The electric potential V at any point O (x, y, z all in metres) in space is given by $V = 4 x^{2} v o l t$ . The electric field at the point $(1 m, 0, 2 m)$ in volt/metre is -

8 along negative x-axis

8 along positive x-axis

16 along negative x-axis

16 along positive z-axis

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Explanation

The electric potential $V (x, y, z) = 4 x^{2} v o l t$

Now $\vec{E} = - (\hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z})$

Now $\frac{\partial V}{\partial x} = 8 x, \frac{\partial V}{\partial y} = 0$ and $\frac{\partial V}{\partial z} = 0$

Hence $\vec{E} = - 8 x \hat{i}$ , so at point (1m, 0, 2m)

$\vec{E} = - 8 \hat{i} v o l t / m e t r e$ or 8 volt/metre along negative X-axis.

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A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre of the sphere is -

0 V

10 V

Same as at point 5 cm away from the surface

Same as at point 25 cm away from the surface

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Explanation

Since potential inside the hollow sphere is same as that on the surface.

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If a unit positive charge is taken from one point to another over an equipotential surface, then -

Work is done on the charge

Work is done by the charge

Work done is constant

No work is done

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Explanation

On the equipotential surface, electric field is normal to the charged surface (where potential exists) so that no work will be done.

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A conductor with a positive charge

Is always at +ve potential

Is always at zero potential

Is always at negative potential

May be at +ve, zero or –ve potential

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Explanation

May be at positive, zero or negative potential, it is according to the way one defines the zero potential.

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Two spheres A and B of radius 4 cm and 6 cm are given charges of 80 μc and 40μc respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is -

20μc from A to B

16μc from A to B

32μc from B to A

32μc from A to B

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Explanation

Total charge $Q = 80 + 40 = 120 μ C$ . By using the formula $Q_{1} ' = Q [\frac{r_{1}}{r_{1} + r_{2}}]$ .

A new charge on sphere A is $Q_{A}^{'} = Q [\frac{r_{A}}{r_{A} + r_{B}}] = 120 [\frac{4}{4 + 6}] = 48 μ C$ .

Initially it was 80 μC i.e., (80-48)= 32 μC charge flows from A to B.

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On rotating a point charge having a charge q around a charge Q in a circle of radius r. The work done will be

$q \times 2 π r$

$\frac{q \times 2 π Q}{r}$

Zero

$\frac{Q}{2 ε_{0} r}$

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Explanation

Since charge Q moving on equipotential surface so work done is zero.

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Two charge +q and –q are situated at a certain distance. At the point exactly midway between them -

Electric field and potential both are zero

Electric field is zero but the potential is not zero

Electric field is not zero but the potential is zero

Neither electric field nor potential is zero

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Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10 C are connected by a copper wire and then they are separated. Then -

Both the spheres will have the same charge of 10 C

Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere

Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere

Surface charge density on the two spheres will be equal

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Explanation

After redistribution, charges on them will be different, but they will acquire common potential

i.e. $k \frac{Q_{1}}{r_{1}} = k \frac{Q_{2}}{r_{2}}$ ⇒ $\frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$

As $σ = \frac{Q}{4 π r^{2}}$ ⇒ $\frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$

⇒ $\frac{σ_{1}}{σ_{2}} = \frac{r_{2}}{r_{1}}$ ⇒ $σ \propto \frac{1}{r}$

i.e. surface charge density on smaller sphere will be more.

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Two equal charges q are placed at a distance of 2a and a third charge –2q is placed at the midpoint. The potential energy of the system is -

$\frac{q^{2}}{8 π ε_{0} a}$

$\frac{6 q^{2}}{8 π ε_{0} a}$

$- \frac{7 q^{2}}{8 π ε_{0} a}$

$\frac{9 q^{2}}{8 π ε_{0} a}$

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Explanation

$U_{s y s t e m} = \frac{1}{4 π ε_{0}} \frac{(q) (- 2 q)}{a} + \frac{1}{4 π ε_{0}} \frac{(- 2 q) (q)}{a} + \frac{1}{4 π ε_{0}} \frac{(q) (q)}{2 a}$

$U_{s y s t e m} = - \frac{7 q^{2}}{8 π ε_{0} a}$

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A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is -

$q E y^{2}$

$q E^{2} y$

$q E y$

$q^{2} E y$

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Explanation

Kinetic energy = Force × Displacement = qEy

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