﻿ Two charges each of equal magnitude 3.2 x 10-19 coulomb but of opposite sign form an electric dipole. The distance between the two charges is 2.4 A∘. If the dipole is placed in an electric field of 5 x 105 volt/metre, then in equilibrium its potential energy will be | Electric Charges and Fields

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Electric Charges and Fields

Multiple Choice Questions

1.

An electron is moving with velocity v on a circular path of radius r in a transverse electric field B. The specific charge (e/m) of the electron is

• Bvr

• $\frac{\mathrm{\nu }}{\mathrm{Br}}$

• Bvr2

• $\frac{\mathrm{B}}{\mathrm{r\nu }}$

2.

Two spheres of same metal have radii a and b. Thet have been connected to a conducting wire. Find the ratio of the electric field intensity upon them.

• a/b

• b/a

• b2 / a

• b2/a2

3.

A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.4 sec. If magnetic flux between the pole pieces is known to be 8 × 10-4 Wb, then induced emf in the wire, is

• 4 × 10-3 V

• 8 × 10-3 V

• 20 × 10-3 V

• 6 × 10-3 V

4.

1 newton/coulomb is equivalent to

• 1 C/V

• 1 J

• 1 V/M

• 1 J/C

5.

The unit of permittivity (${\mathrm{\epsilon }}_{0}$) of space is

• newton-metre/ coulomb2

• coulomb / newton-metre

• coulomb / newton-metre2

• coulomb/ newton-metre2

6.

Length cannot be measure by

• fermi

• micron

• debye

• light year

7.

Using mass (M), length (L), time (T ) and current (A) as fundamental quantities, the dimension of permeability is

• [ M-1 L T-2 A ]

• [ M L2 T-2 A-1 ]

• [ M L T-2A-2 ]

• [ M L T-1 A-1 ]

8.

$\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$

• $\frac{\mathrm{Q}}{6{\mathrm{\epsilon }}_{\mathrm{o}}}$

• $\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{a}}^{2}}$

9.

Figure shows the electric lines of forces emerging from a charged body. If the electric field at A and B are EA and  EB respectively and if the displacement between A and B is r, then

• EA > EB

• EA < EB

• EA = EB/r

• EA = E/ r2

10.Two charges each of equal magnitude 3.2 x 10-19 coulomb but of opposite sign form an electric dipole. The distance between the two charges is 2.4 $\stackrel{\circ }{\mathrm{A}}$. If the dipole is placed in an electric field of 5 x 105 volt/metre, then in equilibrium its potential energy will be3 × 15-23 joule -3.84 × 10-23 joule -6 × 10-23 joule  -2 × 10-26 joule

B.

-3.84 × 10-23 joule

For an electric dipole to be in stable equilibrium, we have

Potential energy U = -pE

= -2qlE

= -q × 2l × E

Here: q = 3.2 × 10-19 coulomb, 2l = 2.4 $\stackrel{\circ }{\mathrm{A}}$

= 2.4 × 10-10 m

E = 5 × 105 volt/metre

$\therefore$     U = 3.2 × 10-19 × 2.4 × 10-10 × 5×105 J

= -3.84 × 10-23 J