﻿ Medical Entrance Exam Question and Answers | Electric Charges and Fields - Zigya

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# Electric Charges and Fields

#### Multiple Choice Questions

1.

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

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

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

B.

Gauss's law tells that the total flux through an arras enclosing a charge Q is  $\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$.

Now as the cube is having six faces and as we can assume a symmetrical distribution of fluxes among its faces, the flux associated with one of its face is  $\frac{\mathrm{Q}}{6{\mathrm{\epsilon }}_{\mathrm{o}}}$.

2.

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 be

• 3 × 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

3.

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

• $\frac{\mathrm{B}}{\mathrm{r\nu }}$

B.

$\frac{\mathrm{\nu }}{\mathrm{Br}}$

If a magnetic field is applied perpendicular to a moving electron, then electron will move on a circular path.

$\therefore$ centripetal force = force on electron due to magnetic field

4.

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

C.

20 × 10-3 V

The charge of flux through the metal wire

Δ$\mathrm{\varphi }$ = 8 × 10-4 Wb

Time taken

Δt = 0.4 sec

∴  Induced e.m.f

e =  $\frac{∆\mathrm{\varphi }}{∆\mathrm{t}}$

=

e = 20 × 10-3 V

5.

Length cannot be measure by

• fermi

• micron

• debye

• light year

C.

debye

A fermi is equivalent to a femtometer and it is older non-SI measurement unit of length.

A micron is a unit of measure in the system equal to 1 millionth of a meter in length ( about 39 millionths of an inch ).

The light-year is a unit of length used to express astronomical distances and measures about 9.46 trillion kilometers ( 9.46 × 1012 km )

Debye is the unit of electric dipole moment. Therefore length cannot be measured by debye.

6.

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

D.

b2/a2

Since both the metal spheres are connected to the same conducting wire, both of them will be having same charge on them.

Let the charge on each of them by q

Then the electric field intensities are given by

Ea$\frac{\mathrm{kq}}{{\mathrm{a}}^{2}}$

Eb$\frac{\mathrm{kq}}{{\mathrm{b}}^{2}}$

∴

7.

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

A.

EA > EB

Since the intensity of the electric lines of force at A is more than that at B, the electric field at A.  EA  >  EB ( electric field at B ).

8.

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

• newton-metre/ coulomb2

• coulomb / newton-metre

• coulomb / newton-metre2

• coulomb/ newton-metre2

D.

coulomb/ newton-metre2

From the relation

F = $\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}$

or  $4{\mathrm{\pi \epsilon }}_{0}$

$\therefore$ Unit of ${\mathrm{\epsilon }}_{0}$ =

=  coulomb/ newton metre2

9.

1 newton/coulomb is equivalent to

• 1 C/V

• 1 J

• 1 V/M

• 1 J/C

C.

1 V/M

From the relation,

Force F = qE

$\therefore$   $\frac{\mathrm{F}}{\mathrm{q}}$ = E = $\frac{\mathrm{V}}{\mathrm{d}}$

10.

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 ]

C.

[ M L T-2A-2 ]

We know that the force per unit length of a wire carrying current due to another parallel wire increase of temperature. Which in turn implies that carrying current is given by

⇒  μo

[ μo ] =

[ μo ] = [ M L T-2 ] [A-2 ]