Free Electrostatics MCQs with Answers

• A. 1/4th of its original value
• B. 1/8th of its original value
• C. 4 times its original value
• D. 8 times its original value

Explanation: The force between two charges is directly proportional to product of magnitude of chargesF =Kq1q2/r2q1'=2q1 & q2'=2q2F' = (2q1)(2q2)/r2F'=…

• A. Zero
• B. Infinite
• C. I-voltm-1
• D. Negative

Explanation: Electric field intensity = kq/r2 r = infinity E = K q / (infinity)2 E = 0 So the electric field intensity becomes zero at infinity.

• A. 1.6
• B. 1.96
• C. 1.986
• D. 1.0006

Explanation: The relative permittivity, or dielectric constant, is a measure of a material's ability to permit electric field lines.

• A. 104 V
• B. 105 V
• C. 106 V
• D. 107 V

Explanation: To calculate the electric potential at a point due to a point charge, we use the formula: V = kQ/r, where V is the potential, k is…

• A. Potential difference
• B. Potential energy
• C. Potential gradient
• D. All of these

Explanation: Potential gradient is the rate of change of potential difference with distance. It is measured in volts per meter (V/m).

• A. Radially inward
• B. Radially outward
• C. Radially zero
• D. None of these

Explanation: A unit positive charge +qo placed in the vicinity of another positive point charge will experience a repulsive force directed radially…

• A. E
• B. E2
• C. 1/E2
• D. E3

Explanation: The correct answer is that the electrostatic energy density is proportional to E2, where E is the electric field intensity.

• A. An imaginary surface
• B. A curved surface
• C. An open surface
• D. A plane surface

Explanation: The correct answer is that a Gaussian surface is an imaginary surface. It is a mathematical construct used in electrostatics to apply…

• A. Electric field
• B. Capacitance
• C. Energy
• D. Charge

Explanation: Option D is correct as The quantity that remains unchanged when a dielectric slab is introduced between the plates of a charged parallel…

• A. 16 x 10^-5 J
• B. 32 x 10^-6 J
• C. 3.2 x 10^-4J
• D. 16 x 10^-4 J

Explanation: The work done in placing a charge of 8 microcoulombs (μC) on a capacitor of capacitance 100 nanofarads (nF) is 3.2×10-4J .A microcoulomb…

• A. The charge will flow from A to B
• B. The charge will flow from B to A
• C. No flow of charge will occur
• D. Both A and B are possible

Explanation: Option A is correct. When metallic spheres A and B, with radii 2m and 4m respectively, carry the same charge and are connected by a copper…

• A. Q only
• B. V only
• C. U only
• D. Q and U only

Explanation: When capacitors are connected in parallel, they share the same potential difference (V) across them, because they are connected directly…

• A. Q/4, 3Q/4
• B. Q/3, 2Q/3
• C. Q/2, Q/2
• D. Q/5, 4Q/5

Explanation: To find the value of charge that should be placed on each sphere so that the force of attraction between them is maximum, we can use the…

• A. C N-1 M-1
• B. watt K-1 mol-1
• C. C2N-1M-2
• D. joule C-1 mol-1

Explanation: Option C is correct because the SI unit of Vacuum Permittivity (ε0) is C2 N-1 M-2 and its value is approximately 8.85 x 10-12.

• A. (1+a)m from 4Q
• B. a/2 m from - Q
• C. 1m from 4Q
• D. Neutral point hence not possible

Explanation: Following is the solution to this question:

• A. 4C
• B. 2C
• C. C/2
• D. C/4

Explanation: Before the metal sheet is inserted, C = εo A/d After the sheet is inserted, the system is equivalent to two capacitors in series, each of…

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Explanation: When a point charge q is placed at a distance r from the center of an uncharged conducting sphere of radius R (where R < r), the potential…

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Explanation: Following is the solution to this question:

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Explanation: Following is the solution to this question in image:

• A. KQ1Q2Q3
• B. 4KQ1Q2
• C. 4KQ2Q3
• D. 2/3 KQ2Q3.

Explanation: Following is the solution to this question in the image: