Electrostatics
Q. What happens to the electric field strength if the distance from a point charge is tripled?
A.
It becomes one-third
B.
It becomes one-ninth
C.
It becomes three times
D.
It remains the same
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Solution
The electric field strength E ∝ 1/r², so if distance is tripled, the field strength becomes 1/9 of the original.
Correct Answer: B — It becomes one-ninth
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Q. What happens to the electric potential as you move away from a positive charge?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes negative
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Solution
The electric potential decreases as you move away from a positive charge.
Correct Answer: B — Decreases
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Q. What happens to the electric potential energy of a charge when it moves against an electric field?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It becomes zero
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Solution
When a charge moves against an electric field, its electric potential energy increases.
Correct Answer: A — It increases
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Q. What happens to the electric potential energy of a system of charges when they are brought closer together?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
The electric potential energy of a system of charges decreases when they are brought closer together, especially if they are of opposite signs.
Correct Answer: B — Decreases
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Q. What happens to the electric potential energy when two like charges are brought closer together?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
The electric potential energy increases as like charges repel each other.
Correct Answer: A — Increases
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Q. What happens to the energy stored in a capacitor if the voltage across it is doubled?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
The energy stored in a capacitor is given by U = 1/2 C V². If the voltage is doubled, the energy increases by a factor of four.
Correct Answer: B — It quadruples
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Q. What happens to the potential difference across a capacitor when it is fully charged?
A.
It becomes zero
B.
It becomes maximum
C.
It becomes minimum
D.
It fluctuates
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Solution
When a capacitor is fully charged, the potential difference across its plates becomes maximum and remains constant until it is discharged.
Correct Answer: B — It becomes maximum
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Q. What happens to the voltage across a capacitor when it is fully charged?
A.
It becomes zero
B.
It equals the supply voltage
C.
It becomes negative
D.
It fluctuates
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Solution
When a capacitor is fully charged, the voltage across it equals the supply voltage.
Correct Answer: B — It equals the supply voltage
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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 relative permittivity 5?
A.
5.5 × 10^-11 F
B.
5.5 × 10^-10 F
C.
5.5 × 10^-9 F
D.
5.5 × 10^-8 F
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Solution
C = ε₀ * ε_r * A / d = (8.85 × 10^-12 F/m) * 5 * (0.01 m²) / (0.001 m) = 4.425 × 10^-10 F.
Correct Answer: B — 5.5 × 10^-10 F
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Q. What is the capacitance of a parallel plate capacitor with area 0.1 m² and separation 0.01 m filled with air (ε₀ = 8.85 × 10^-12 F/m)?
A.
8.85 × 10^-12 F
B.
8.85 × 10^-10 F
C.
8.85 × 10^-9 F
D.
8.85 × 10^-8 F
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Solution
Capacitance C = ε₀ * A / d = (8.85 × 10^-12 F/m) * (0.1 m²) / (0.01 m) = 8.85 × 10^-10 F.
Correct Answer: B — 8.85 × 10^-10 F
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Q. What is the capacitance of a parallel plate capacitor with area A and separation d?
A.
ε₀ * A / d
B.
A / (ε₀ * d)
C.
d / (ε₀ * A)
D.
ε₀ * d / A
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Solution
The capacitance C of a parallel plate capacitor is given by the formula C = ε₀ * A / d, where ε₀ is the permittivity of free space.
Correct Answer: A — ε₀ * A / d
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Q. What is the capacitance of a parallel plate capacitor with plate area A and separation d?
A.
ε₀A/d
B.
d/ε₀A
C.
A/ε₀d
D.
ε₀d/A
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Solution
The capacitance C of a parallel plate capacitor is given by the formula C = ε₀A/d, where ε₀ is the permittivity of free space.
Correct Answer: A — ε₀A/d
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Q. What is the direction of the electric field due to a negative point charge?
A.
Away from the charge
B.
Towards the charge
C.
Perpendicular to the charge
D.
None of the above
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Solution
The electric field due to a negative charge points towards the charge.
Correct Answer: B — Towards the charge
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Q. What is the direction of the electric field due to a positive charge?
A.
Towards the charge
B.
Away from the charge
C.
Perpendicular to the charge
D.
None of the above
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Solution
The electric field due to a positive charge points away from the charge.
Correct Answer: B — Away from the charge
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Q. What is the direction of the electric field due to a positive point charge?
A.
Towards the charge
B.
Away from the charge
C.
Perpendicular to the charge
D.
None of the above
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Solution
The electric field lines point away from a positive charge.
Correct Answer: B — Away from the charge
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Q. What is the effect of a dielectric material on the electric field between the plates of a parallel plate capacitor?
A.
Increases the electric field
B.
Decreases the electric field
C.
Has no effect
D.
Reverses the electric field direction
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Solution
Inserting a dielectric material between the plates of a capacitor decreases the electric field due to polarization.
Correct Answer: B — Decreases the electric field
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Q. What is the effect of a dielectric material on the electric field inside a capacitor?
A.
Increases the electric field
B.
Decreases the electric field
C.
Has no effect
D.
Reverses the electric field direction
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Solution
Inserting a dielectric material into a capacitor decreases the electric field inside the capacitor due to polarization.
Correct Answer: B — Decreases the electric field
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Q. What is the effect of dielectric material on capacitance?
A.
Increases capacitance
B.
Decreases capacitance
C.
No effect
D.
Depends on the charge
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Solution
The presence of a dielectric material increases the capacitance of a capacitor.
Correct Answer: A — Increases capacitance
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Q. What is the effect of dielectric material on the capacitance of a capacitor?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
The presence of a dielectric material increases the capacitance of a capacitor.
Correct Answer: A — Increases
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Q. What is the effect of increasing the permittivity of the medium on the electric field due to a point charge?
A.
Electric field increases
B.
Electric field decreases
C.
Electric field remains the same
D.
Electric field becomes zero
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Solution
Increasing the permittivity of the medium decreases the electric field due to a point charge, as E = Q/(4πε₀r²).
Correct Answer: B — Electric field decreases
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Q. What is the effect of increasing the plate area of a capacitor on its capacitance?
A.
Capacitance increases
B.
Capacitance decreases
C.
Capacitance remains the same
D.
Capacitance becomes zero
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Solution
Increasing the plate area A of a capacitor increases its capacitance, as C is directly proportional to A.
Correct Answer: A — Capacitance increases
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Q. What is the effect of increasing the plate area of a parallel plate capacitor?
A.
Capacitance decreases
B.
Capacitance increases
C.
Capacitance remains the same
D.
Capacitance becomes zero
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Solution
Increasing the plate area A of a parallel plate capacitor increases its capacitance, as C = ε₀ * A / d.
Correct Answer: B — Capacitance increases
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Q. What is the effect of increasing the radius of a spherical Gaussian surface enclosing a fixed charge?
A.
Electric field increases.
B.
Electric field decreases.
C.
Electric field remains constant.
D.
Electric field becomes zero.
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Solution
The electric field due to a point charge decreases with distance, but the total flux remains constant as the charge is fixed.
Correct Answer: C — Electric field remains constant.
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Q. What is the effect of increasing the surface charge density on the electric field of a charged plane sheet?
A.
Increases
B.
Decreases
C.
No effect
D.
Becomes zero
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Solution
The electric field due to a charged plane sheet is directly proportional to the surface charge density; increasing σ increases E.
Correct Answer: A — Increases
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Q. What is the effect of inserting a dielectric material between the plates of a capacitor?
A.
Increases capacitance
B.
Decreases capacitance
C.
No effect on capacitance
D.
Changes the voltage across the plates
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Solution
Inserting a dielectric material increases the capacitance of the capacitor by a factor equal to the dielectric constant of the material.
Correct Answer: A — Increases capacitance
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Q. What is the effect of temperature on the capacitance of a capacitor?
A.
Increases capacitance
B.
Decreases capacitance
C.
No effect
D.
Depends on the dielectric
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Solution
The effect of temperature on capacitance depends on the dielectric material used in the capacitor.
Correct Answer: D — Depends on the dielectric
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Q. What is the electric field at a distance d from an infinitely long charged wire with linear charge density λ?
A.
λ/(2πε₀d)
B.
λ/(4πε₀d²)
C.
λ/(2πε₀d²)
D.
0
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Solution
The electric field due to an infinitely long charged wire is given by E = λ/(2πε₀d).
Correct Answer: A — λ/(2πε₀d)
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Q. What is the electric field at a distance of 1m from a charge of +1μC?
A.
9 × 10^9 N/C
B.
9 × 10^6 N/C
C.
9 × 10^3 N/C
D.
9 × 10^12 N/C
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Solution
Electric field E = k * q / r^2 = (9 × 10^9) * (1 × 10^-6) / (1)^2 = 9 × 10^3 N/C.
Correct Answer: B — 9 × 10^6 N/C
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Q. What is the electric field at a distance r from a uniformly charged disk of radius R and surface charge density σ?
A.
σ/(2ε₀)
B.
σ/(4ε₀)
C.
σ/(2ε₀) * (1 - r/√(R² + r²))
D.
Zero
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Solution
The electric field at a distance r from a uniformly charged disk is given by E = σ/(2ε₀) * (1 - r/√(R² + r²)).
Correct Answer: C — σ/(2ε₀) * (1 - r/√(R² + r²))
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Q. What is the electric field at a distance r from a uniformly charged sphere of radius R and total charge Q, when r > R?
A.
Q/(4πε₀r²)
B.
Q/(4πε₀R²)
C.
Zero
D.
Q/(4πε₀R)
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Solution
For r > R, the electric field behaves as if all the charge were concentrated at the center, thus E = Q/(4πε₀r²).
Correct Answer: A — Q/(4πε₀r²)
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