Electrostatics & Circuits Study for Competitive Exams

Electrostatics & Circuits

Coulombs Law and Electric Field Coulombs Law and Electric Field - Capacitance and Dielectrics Coulombs Law and Electric Field - Circuit Analysis Techniques Coulombs Law and Electric Field - Electric Field Problems Coulombs Law and Electric Field - Transient Response in RC Circuits Current, Resistance and Ohms Law Current, Resistance and Ohms Law - Capacitance and Dielectrics Current, Resistance and Ohms Law - Circuit Analysis Techniques Current, Resistance and Ohms Law - Electric Field Problems Current, Resistance and Ohms Law - Transient Response in RC Circuits DC Circuits and Kirchhoffs Laws DC Circuits and Kirchhoffs Laws - Capacitance and Dielectrics DC Circuits and Kirchhoffs Laws - Circuit Analysis Techniques DC Circuits and Kirchhoffs Laws - Electric Field Problems DC Circuits and Kirchhoffs Laws - Transient Response in RC Circuits Electric Potential and Capacitance Electric Potential and Capacitance - Capacitance and Dielectrics Electric Potential and Capacitance - Circuit Analysis Techniques Electric Potential and Capacitance - Electric Field Problems Electric Potential and Capacitance - Transient Response in RC Circuits Magnetic Fields and Electromagnetic Induction Magnetic Fields and Electromagnetic Induction - Capacitance and Dielectrics Magnetic Fields and Electromagnetic Induction - Circuit Analysis Techniques Magnetic Fields and Electromagnetic Induction - Electric Field Problems Magnetic Fields and Electromagnetic Induction - Transient Response in RC Circuits

Q. What is the capacitance of a capacitor with a charge of 5 µC and a potential difference of 10 V?

A. 0.5 µF
B. 1 µF
C. 2 µF
D. 0.2 µF

Solution

Capacitance (C) is given by C = Q/V. Here, C = 5 µC / 10 V = 0.5 µF.

Correct Answer: A — 0.5 µF

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Q. What is the capacitance of a capacitor with a charge of 6 µC and a voltage of 12 V?

A. 0.5 µF
B. 0.25 µF
C. 0.75 µF
D. 1 µF

Solution

C = Q/V = 6 x 10^-6 C / 12 V = 0.5 x 10^-6 F = 0.5 µF.

Correct Answer: A — 0.5 µF

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Q. What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of k=2?

A. 5.5 µF
B. 4.4 µF
C. 2.2 µF
D. 1.1 µF

Solution

C = k * ε₀ * (A/d) = 2 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m) = 4.4 µF.

Correct Answer: B — 4.4 µF

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Q. What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of k=5?

A. 5.5 µF
B. 5.0 µF
C. 4.5 µF
D. 6.0 µF

Solution

C = k * ε0 * (A/d) = 5 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m) = 5.0 µF.

Correct Answer: B — 5.0 µF

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Q. What is the capacitance of a parallel plate capacitor with an area of 2 m² and a separation of 0.01 m filled with a dielectric of relative permittivity 5?

A. 0.88 µF
B. 1.77 µF
C. 2.65 µF
D. 3.54 µF

Solution

Capacitance C = (ε₀ * ε_r * A) / d = (8.85 x 10^-12 F/m * 5 * 2 m²) / 0.01 m = 8.85 x 10^-11 F = 1.77 µF.

Correct Answer: B — 1.77 µF

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Q. What is the charge stored in a capacitor of 5 µF at a voltage of 10 V?

A. 50 µC
B. 100 µC
C. 25 µC
D. 75 µC

Solution

Charge Q = C * V = 5 x 10^-6 F * 10 V = 50 µC.

Correct Answer: A — 50 µC

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Q. What is the charge stored in a capacitor of 5 µF when connected to a 10 V battery?

A. 0.05 mC
B. 0.1 mC
C. 0.2 mC
D. 0.5 mC

Solution

Charge Q = C * V = 5 x 10^-6 F * 10 V = 0.05 mC.

Correct Answer: B — 0.1 mC

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Q. What is the current flowing through a 20 Ω resistor when a voltage of 100 V is applied?

A. 5 A
B. 10 A
C. 15 A
D. 20 A

Solution

Using Ohm's Law: I = V / R = 100 V / 20 Ω = 5 A.

Correct Answer: B — 10 A

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Q. What is the current through a 10 Ω resistor if the voltage across it is 30 V?

A. 3 A
B. 1 A
C. 5 A
D. 2 A

Solution

Using Ohm's Law: I = V/R = 30 V / 10 Ω = 3 A.

Correct Answer: A — 3 A

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Q. What is the current through a 12Ω resistor connected to a 36V source?

A. 2A
B. 3A
C. 4A
D. 5A

Solution

Using Ohm's law, I = V/R = 36V / 12Ω = 3A.

Correct Answer: B — 3A

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Q. What is the current through a 15 ohm resistor connected to a 45V source?

A. 3 A
B. 2 A
C. 1 A
D. 4 A

Solution

Using Ohm's Law, I = V/R. Here, I = 45V / 15Ω = 3 A.

Correct Answer: A — 3 A

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Q. What is the current through a 5 Ω resistor connected to a 15 V battery?

A. 3 A
B. 2 A
C. 1 A
D. 0.5 A

Solution

Using Ohm's law, I = V/R = 15 V / 5 Ω = 3 A.

Correct Answer: B — 2 A

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Q. What is the current through a 5Ω resistor if the voltage across it is 15V?

A. 3 A
B. 2 A
C. 1 A
D. 4 A

Solution

Using Ohm's law, I = V/R = 15V / 5Ω = 3 A.

Correct Answer: A — 3 A

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Q. What is the dielectric constant of a material if a capacitor with air as dielectric has a capacitance of 10 µF and the same capacitor with the material has a capacitance of 30 µF?

A. 1
B. 2
C. 3
D. 4

Solution

Dielectric constant ε_r = C_material / C_air = 30 µF / 10 µF = 3.

Correct Answer: C — 3

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Q. What is the effect of adding a dielectric to a capacitor?

A. Increases capacitance
B. Decreases capacitance
C. No effect on capacitance
D. Increases resistance

Solution

Adding a dielectric increases the capacitance of the capacitor.

Correct Answer: A — Increases capacitance

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Q. What is the effect of connecting a capacitor in parallel with a resistor in a DC circuit?

A. Increases current
B. Decreases current
C. Stores energy
D. No effect

Solution

A capacitor stores energy when connected in parallel with a resistor, affecting the circuit's behavior over time.

Correct Answer: C — Stores energy

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Q. What is the effect of increasing the capacitance in an RC circuit on the time constant?

A. Time constant decreases
B. Time constant increases
C. Time constant remains the same
D. Time constant becomes zero

Solution

Increasing the capacitance in an RC circuit increases the time constant τ = R*C, making the charging and discharging processes slower.

Correct Answer: B — Time constant increases

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Q. What is the effect of increasing the distance between the plates of a parallel plate capacitor?

A. Capacitance increases
B. Capacitance decreases
C. Capacitance remains the same
D. Capacitance becomes zero

Solution

Capacitance decreases as the distance between the plates increases.

Correct Answer: B — Capacitance decreases

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Q. What is the effect of increasing the distance between the plates of a parallel plate capacitor on its capacitance?

A. Capacitance increases
B. Capacitance decreases
C. Capacitance remains the same
D. Capacitance becomes zero

Solution

Capacitance (C) is inversely proportional to the distance (d) between the plates. Increasing d decreases C.

Correct Answer: B — Capacitance decreases

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Q. What is the effect of increasing the resistance (R) in an RC charging circuit on the time constant (τ)?

A. τ increases
B. τ decreases
C. τ remains the same
D. τ becomes zero

Solution

Increasing the resistance in an RC circuit increases the time constant τ, since τ = R * C.

Correct Answer: A — τ increases

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Q. What is the effect of increasing the resistance in an RC circuit on the time constant?

A. It decreases the time constant
B. It has no effect
C. It increases the time constant
D. It doubles the time constant

Solution

Increasing the resistance in an RC circuit increases the time constant, as τ = R * C.

Correct Answer: C — It increases the time constant

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Q. What is the electric field (E) at a distance of 2 meters from a point charge of 5 microcoulombs?

A. 1125 N/C
B. 450 N/C
C. 225 N/C
D. 900 N/C

Solution

Using Coulomb's law, E = k * |q| / r^2, where k = 8.99 x 10^9 N m^2/C^2, q = 5 x 10^-6 C, and r = 2 m. E = (8.99 x 10^9) * (5 x 10^-6) / (2^2) = 1125 N/C.

Correct Answer: A — 1125 N/C

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Q. What is the electric field (E) at a distance of 2m from a point charge of 10μC?

A. 2250 N/C
B. 500 N/C
C. 4500 N/C
D. 1000 N/C

Solution

Using Coulomb's law, E = k * |Q| / r^2 = (8.99 x 10^9 N m²/C²) * (10 x 10^-6 C) / (2 m)^2 = 2247.5 N/C, approximately 2250 N/C.

Correct Answer: A — 2250 N/C

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Q. What is the electric field (E) due to a point charge (Q) at a distance (r)?

A. E = k * Q / r^2
B. E = k * Q * r^2
C. E = Q / (4 * π * ε * r)
D. E = Q / (4 * π * ε * r^2)

Solution

The electric field due to a point charge is given by E = k * Q / r^2.

Correct Answer: A — E = k * Q / r^2

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Q. What is the electric field (E) due to a uniformly charged infinite plane sheet with surface charge density (σ)?

A. E = σ / (2 * ε)
B. E = σ / ε
C. E = σ / (4 * π * ε)
D. E = 0

Solution

The electric field due to a uniformly charged infinite plane sheet is E = σ / (2 * ε).

Correct Answer: A — E = σ / (2 * ε)

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Q. What is the electric field at a distance of 2 m from a point charge of +5 µC?

A. 0.56 N/C
B. 1.12 N/C
C. 2.24 N/C
D. 4.48 N/C

Solution

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (2 m)^2 = 1.12 N/C.

Correct Answer: C — 2.24 N/C

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Q. What is the electric field between two parallel plates separated by 0.1 m with a potential difference of 100 V?

A. 1000 N/C
B. 500 N/C
C. 100 N/C
D. 10 N/C

Solution

E = V/d = 100 V / 0.1 m = 1000 N/C.

Correct Answer: A — 1000 N/C

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Q. What is the electric field due to a point charge of -4 µC at a distance of 0.25 m?

A. -5760 N/C
B. -1440 N/C
C. 5760 N/C
D. 1440 N/C

Solution

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (4 x 10^-6 C) / (0.25 m)^2 = -5760 N/C.

Correct Answer: A — -5760 N/C

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Q. What is the electric field inside a parallel plate capacitor with a potential difference of 100 V and plate separation of 0.01 m?

A. 10000 N/C
B. 1000 N/C
C. 100 N/C
D. 10 N/C

Solution

E = V/d = 100 V / 0.01 m = 10000 N/C.

Correct Answer: A — 10000 N/C

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Q. What is the electric field strength at a distance of 0.2 m from a point charge of +5 µC?

A. 112.5 N/C
B. 112.5 kN/C
C. 25 N/C
D. 25 kN/C

Solution

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (0.2 m)^2 = 112.5 N/C.

Correct Answer: A — 112.5 N/C

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Electrostatics & Circuits

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