Electrostatics Test

Result

Electrostatics Test - 16

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Question 1
4 / -1

The electric field intensity E, due to an electric dipole of moment p, at a point on the equatorial line is

A
parallel to the axis of the dipole and opposite to the direction of the dipole moment p

B
perpendicular to the axis of the dipole and is directed away from it

C
parallel to the dipole moment

D
perpendicular to the axis of the dipole and is directed towards it

Solution
Question 2
4 / -1

The electric potential due to a small electric dipole at a large distance r from the centre of the dipole is proportional to

A

B

C

D

Solution
Question 3
4 / -1

An electric dipole is placed at an angle of 30° to a non–uniform electric field. The dipole will experience

A
a translational force only in the direction of the field

B
a translational force only in a direction normal to the direction of the field

C
a torque as well as a translational force

D
a torque only

Solution
Question 4
4 / -1

Electric field strength due to a dipole at a point on the axial line of dipole is

A
from positive charge to negative charge

B
from negative charge to positive charge

C
along the equatorial line

D
at an angle to axial line

Solution
Question 5
4 / -1

The electric field due to an electric dipole at a distance r from its centre in axial position is E. If the dipole is rotated through an angle of 90° about its perpendicular axis, the electric field at the same point will be

A
E

B
E / 4

C
E / 2

D
2E

Solution
Question 6
4 / -1

An electric dipole of moment p is placed at the origin along the x-axis. The electric field at a point P, whose position vector makes an angle θ with the x-axis, will make an angle with the x–axis where,

A
α

B
θ

C
θ + α

D
2 θ + α

Solution
Question 7
4 / -1

A

B

C

D

Solution
Question 8
4 / -1

Two non-conducting spheres of radii R₁ and R₂ and carrying uniform volume charge densities + ρ and –ρ, respectively, are placed such that they partially overlap, as shown in figure. At all points in the overlapping region

A
the electrostatic field is zero

B
the electrostatic potential is constant

C
the electrostatic field is constant in magnitude

D
the electrostatic field has not same direction

Solution
Question 9
4 / -1

A charge 10 μC is placed at the centre of a hemisphere of radius R = 10 cm as shown. The electric flux through the hemisphere (in MKS units) is

A
20 × 10⁵

B
10 × 10⁵

C
6 × 10⁵

D
2 × 10⁵

Solution
Question 10
4 / -1

The electrostatic potential inside a charged spherical ball is given by ϕ = ar² + b where r is the distance from the centre a,and b are constants. Then, the charge density inside the ball is

A
–6 a ε₀ r

B
–24 π a ε₀

C
–6 a ε₀

D
–24 π a ε₀ r

Solution