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Test: Electric Charges and Fields - 1 - CUET Humanities MCQ


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Test: Electric Charges and Fields - 1 - Question 1

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if Q is equal to

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 1

Since the hypotenuse of the triangle is √2 a, the net electrostatic energy is:

For U = 0, we require

Or, 
Hence, option 2 is the correct answer.

Test: Electric Charges and Fields - 1 - Question 2

When a body is negatively charged by friction, it means the body has

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 2

The body acquired a negative charge either by gaining electron or losing proton, losing proton is not easy for any body because it is only possible in nuclear reaction, So Acquired excess of electron is the correct option.

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Test: Electric Charges and Fields - 1 - Question 3

Two point charges 3 × 10-6 C and 8 × 10-6 C repel each other by a force of 6 × 10-3 N. If each of them is given an additional charge of -6 × 10-6 C, then the force between them will be

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 3

Key Idea: Like charges repel each other while unlike charges attract each other.
From Coulomb's law, the force of attraction/repulsion between two point charges q1 and q2 placed a distance r apart is given by
F =  N
When similar charges are taken
q1 = 3 × 10-6 C, q2 = 8 × 10-6 C
F =  ...(i)
(repulsive)
When additonal charge - 6 × 106 C is given to each charge, then
F' = 
(attractive)
∴ F' =  N ... (ii)
Dividing Eq. (ii) by Eq. (i), we get

⇒ F' = - F/4
= -  = - 1.5 × 10-3 N
Negative sign indicates, force is attractive.

Test: Electric Charges and Fields - 1 - Question 4

The process of acquiring temporary electrification under the influence of a charged body is called

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 4

The process of acquiring temporary electrification under the influence of a charged body is called induction.

Test: Electric Charges and Fields - 1 - Question 5

A pendulum bob of mass 80 milligrams, which is carrying a charge of 2 x 10-8 C, is at rest in a uniform horizontal electric field of 20,000 V/m. What angle (θ) will it make with the vertical?

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 5


At equilibrium,
T cos θ = mg --- (1)
T sinθ = qE --- ( 2)
Dividing (2) by (1), we get:


Test: Electric Charges and Fields - 1 - Question 6

The angle between electric dipole moment and electric field intensity due to an electric dipole at a point on its equatorial line is

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 6

The angle between electric dipole moment and electric field intensity due to an electric dipole at a point on its equatorial line is 180° as both of them are in opposite directions.

Test: Electric Charges and Fields - 1 - Question 7

An electric dipole placed in a non-uniform electric field experiences

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 7

The correct option is (3). In a non-uniform electric field, a dipole experiences a force, which gives it a translational motion, and a torque, which gives it a rotational motion.

Test: Electric Charges and Fields - 1 - Question 8

Three charges 1 μC, 1 μC and 2 μC are respectively kept at the vertices A, B and C of an equilateral triangle ABC of side 10 cm. The resultant force on the charge at C is

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 8


Here, FCA = FCB = FC
Hence, the resultant force on the charge at C is


Test: Electric Charges and Fields - 1 - Question 9

A metallic solid sphere is placed in a uniform electric field as shown. Which path will the lines of force follow?

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 9

The electric field is always perpendicular to the surface of a conductor. On the surface of a metallic solid sphere, the electric field is perpendicular to the surface and directed towards the centre of the sphere. Hence, the correct option is (4).

Test: Electric Charges and Fields - 1 - Question 10

Three infinite long plane sheets carrying uniform charge densities σ1 = -σ, σ​​​​​​​2 = +2σ and σ​​​​​​​3 = +3σ​​​​​​​ are placed parallel to the x-z plane at y = a, y = 3a and y = 4a as shown in the figure. The electric field at point P is

Detailed Solution for Test: Electric Charges and Fields - 1 - Question 10

The electric field at a point P due to an infinite long plane sheet carrying a uniform charge density σ is given by .
It is independent of the distance of point P from the sheet and is therefore, uniform.
The direction of the electric field is away from the sheet and perpendicular to it, if σ is positive and is towards the sheet and perpendicular to it, if σ is negative.
Hence, 

and, 
From the superposition principle, the net electric field at point is E = E1 + E2 + E

 which is option (3).

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