  Difficult Questions

## Electric Charges And Fields

- Are you well prepared? Practice hard questions to be more confident about the chapter
1
When two charged objects immersed in a dielectric liquid then the number of forces along with electrostatic force will come in picture. Let's practice this question.

Two charged balls of the same radius and weight suspended on threads of equal length are immersed into a liquid having density of and a dielectric constant . The density of the material of the balls for the angles of divergence of the threads in the air and in the dielectric to be the same is:

A
B
C
D
2
A charge can do SHM in between two opposite charges after small displacement. Figure out the situation when this charge will do SHM.
A charge +q is placed at and another charge is placed at . A charge is placed at the origin. If it is slightly displaced along y axis. Then :
A
it will move away
B
it will oscillate but will not execute SHM
C
it will execute SHM
D
it will stand at the displaced position
3
Electrostatic interaction between dipole and charged rod are quite complex. Force on the rod due to dipole can be calculated by integrating the value of force on a small length of rod 'dl' having charge .
A small electric dipole having dipole moment is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod  placed along x axis, with one end coinciding with origin. If linear charge density of rod is and distance of dipole from rod is 'a', then calculate the electric force acting on dipole.
A
B
C
D
4
A charged block is maintaining equilibrium on a smooth inclined surface in presence of the horizontal electric field. In this situation, three different forces will act on the block at different-different angles. Let's practice this question and find the equilibrium condition and the direction of friction.
A particle of mass 1 Kg carrying 0.01C charge is at rest on an inclined plane of angle with the horizontal when an electric field of is applied parallel to the horizontal as shown. The coefficient of friction is :
A
B
C
D
5
Let's practice, the concepts of projectile motion of an electron in presence of an electric field.
An electron is projected, as shown in the figure, at a speed of , at an angle of . It is given that is directed upward, and . The electron will strike at :
A
Upper plate
B
Mid point of the lower plate
C
Opposite end of the lower plate
D
No where
6
A charged particle acquires some velocity from rest in presence of the electric field. Otherhand, In direction of the electric field, One surface is placed. Let's think how much flux will link with the given surface and solve the below question.

A particle that carries a charge is placed at rest in uniform electric field . It experiences a force and moves in a certain time t, it is observed to acquire a velocity m/s. The given electric field intersects a surface of area in the X -Z plane. Electric flux through surface is:

A
B
C
D
7
Calculate the charge enclosed by the closed surface when it is kept in a linearly varying electric field.
The electric field in a region is . The charge contained inside a cubical volume bounded by the surfaces is (where x, y, z are in cm) :
A
B
C
D
8
Find the condition of zero electric fields outside the ball when a ball has a fixed charge on its surface but in interior volume, charge distribution is varying with radial distance.
A system consists of a ball of radius carrying a uniformly distributed charge and surrounding space filled with a charge of volume density where is a constant and is the distance from the centre of the ball. Find the charge on the ball for which the magnitude of electric field strength outside the ball is constant ? The dielectric constant of the ball and the surrounding may be taken equal to unity.
9
A charged particle is applying force 'F' on the uniformly charged disc, then find the amount of electric flux linkage with the disc.
A point charge placed on the axis of a uniformly charged disc experiences a force F due to the disc. If the surface charge density on the disc is , the electric flux through the disc due to the point charge will be :
A
B
C
D