Electrostatics
Q. If a charge of +Q is uniformly distributed over a spherical shell of radius R, what is the electric field inside the shell?
A.
0
B.
Q/4πε₀R²
C.
Q/ε₀R²
D.
Q/4πε₀
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Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer: A — 0
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Q. If a charge of +Q is uniformly distributed over a spherical shell, what is the electric field inside the shell?
A.
0
B.
Q/4πε₀r²
C.
Q/ε₀
D.
Q/4πε₀
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Solution
Inside a uniformly charged spherical shell, the electric field is zero due to symmetry.
Correct Answer: A — 0
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Q. If a charge Q is placed at one corner of a cube, what is the electric flux through one face of the cube?
A.
Q/6ε₀
B.
Q/3ε₀
C.
Q/4ε₀
D.
Q/12ε₀
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Solution
By symmetry, the flux through one face of the cube is Q/6ε₀, as the charge is shared equally among the six faces.
Correct Answer: A — Q/6ε₀
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Q. If a charge Q is uniformly distributed over a sphere of radius R, what is the electric field at a distance r from the center where r > R?
A.
Q/(4πε₀r²)
B.
Q/(4πε₀R²)
C.
0
D.
Q/(4πε₀R²) * (R/r)²
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Solution
For r > R, the electric field behaves as if all the charge were concentrated at the center, given by E = Q/(4πε₀r²).
Correct Answer: A — Q/(4πε₀r²)
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Q. If a charge Q is uniformly distributed over a spherical surface of radius R, what is the electric field at a point inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
D.
Q/(4πε₀R³)
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Solution
Inside a uniformly charged spherical shell, the electric field is zero.
Correct Answer: B — 0
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Q. If a charge Q is uniformly distributed over a spherical surface of radius R, what is the electric field at a point outside the sphere at a distance r from the center (r > R)?
A.
0
B.
Q/(4πε₀r²)
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
For a point outside the sphere, the electric field behaves as if all the charge were concentrated at the center, hence E = Q/(4πε₀r²).
Correct Answer: B — Q/(4πε₀r²)
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Q. If a point charge Q is placed at the center of a spherical Gaussian surface of radius R, what is the total electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀R²
D.
Q/4πε₀
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Solution
The total electric flux through the Gaussian surface is given by Φ = Q/ε₀, where Q is the charge enclosed.
Correct Answer: B — Q/ε₀
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Q. If a point charge Q is placed at the center of a spherical Gaussian surface of radius R, what is the electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/2ε₀
D.
Q/4ε₀
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Solution
The electric flux through a closed surface is given by Φ = Q_enc/ε₀. Here, Q_enc = Q, so Φ = Q/ε₀.
Correct Answer: B — Q/ε₀
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Q. If a point charge Q is placed at the center of a spherical Gaussian surface, what is the total electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀
D.
4πQ/ε₀
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Solution
According to Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed divided by ε₀, thus Φ = Q/ε₀.
Correct Answer: B — Q/ε₀
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Q. If the charge density of a non-conducting sphere increases linearly with radius, how does the electric field vary inside the sphere?
A.
Linearly with radius
B.
Quadratically with radius
C.
Constant
D.
Inversely with radius
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Solution
The electric field inside a non-conducting sphere with linearly increasing charge density varies linearly with radius.
Correct Answer: A — Linearly with radius
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Q. If the charge density of a non-uniform spherical charge distribution varies as ρ(r) = kr², what is the electric field at the center of the sphere?
A.
0
B.
k/3ε₀
C.
k/4ε₀
D.
k/2ε₀
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Solution
At the center of a non-uniform spherical charge distribution, the electric field is zero due to symmetry.
Correct Answer: A — 0
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Q. If the charge density of a spherical charge distribution increases linearly with radius, how does the electric field vary inside the sphere?
A.
Linearly with radius
B.
Quadratically with radius
C.
Inversely with radius
D.
Constant
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Solution
The electric field inside a non-uniform charge distribution varies quadratically with radius due to the increasing charge density.
Correct Answer: B — Quadratically with radius
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Q. If the charge inside a closed surface is doubled, what happens to the electric flux through the surface?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
According to Gauss's law, the electric flux is directly proportional to the enclosed charge, so it doubles.
Correct Answer: A — It doubles
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Q. If the charge on a capacitor is doubled while the voltage remains constant, what happens to the capacitance?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance C = Q/V; if Q is doubled and V remains constant, C must halve.
Correct Answer: B — It halves
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Q. If the charge on a capacitor is doubled, what happens to the energy stored in it?
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 * V^2. If the charge is doubled, the energy stored quadruples.
Correct Answer: B — It quadruples
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Q. If the distance between the plates of a capacitor is doubled, what happens to its capacitance?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance is inversely proportional to the distance between the plates. If the distance d is doubled, the capacitance C becomes C/2.
Correct Answer: B — It halves
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Q. If the distance between the plates of a capacitor is halved, what happens to its capacitance?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance is inversely proportional to the distance between the plates. If the distance is halved, the capacitance doubles.
Correct Answer: A — It doubles
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Q. If the distance between two charges is doubled, how does the electrostatic force between them change?
A.
It doubles
B.
It halves
C.
It quadruples
D.
It reduces to one-fourth
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Solution
According to Coulomb's law, F ∝ 1/r^2. If r is doubled, F becomes F/4.
Correct Answer: D — It reduces to one-fourth
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Q. If the distance between two charges is doubled, how does the force between them change?
A.
It doubles
B.
It halves
C.
It quadruples
D.
It reduces to one-fourth
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Solution
According to Coulomb's law, force is inversely proportional to the square of the distance. If the distance is doubled, the force reduces to one-fourth.
Correct Answer: D — It reduces to one-fourth
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Q. If the distance between two charges is halved, how does the force between them change?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
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Solution
According to Coulomb's law, the force F between two charges is inversely proportional to the square of the distance. If the distance is halved, the force becomes four times greater.
Correct Answer: B — It quadruples
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Q. If the distance between two charges is halved, what happens to the force between them?
A.
It doubles
B.
It halves
C.
It quadruples
D.
It remains the same
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Solution
According to Coulomb's law, F ∝ 1/r^2; halving the distance increases the force by a factor of 4.
Correct Answer: C — It quadruples
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Q. If the electric field at a point is 100 N/C and the charge at that point is +5μC, what is the force experienced by the charge?
A.
0.5 N
B.
0.2 N
C.
0.1 N
D.
0.4 N
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Solution
Force F = E * q = 100 N/C * 5 × 10^-6 C = 0.0005 N = 0.5 N.
Correct Answer: A — 0.5 N
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Q. If the electric field at a point is 200 N/C and the charge at that point is +5μC, what is the force experienced by the charge?
A.
0.1 N
B.
1 N
C.
0.5 N
D.
2 N
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Solution
F = E * q = 200 N/C * 5 × 10^-6 C = 1 N.
Correct Answer: B — 1 N
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Q. If the electric field at a point is 200 N/C directed towards the charge, what is the nature of the charge?
A.
Positive
B.
Negative
C.
Neutral
D.
Cannot be determined
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Solution
The electric field is directed towards the charge, indicating that the charge is negative.
Correct Answer: B — Negative
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Q. If the electric field at a point is 200 N/C directed towards the positive x-axis, what is the potential difference between two points A and B separated by 3m along the x-axis?
A.
600 V
B.
400 V
C.
200 V
D.
0 V
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Solution
V = -E * d = -200 N/C * 3m = -600 V. The potential at B is 600 V lower than at A.
Correct Answer: B — 400 V
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Q. If the electric field at a point is zero, what can be said about the charges producing it?
A.
There are no charges nearby
B.
The charges are equal and opposite
C.
The charges are all positive
D.
The charges are all negative
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Solution
The electric field is zero when the vector sum of the electric fields due to all charges is zero, which occurs when there are equal and opposite charges.
Correct Answer: B — The charges are equal and opposite
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Q. If the electric field due to a charged infinite plane sheet is E, what is the electric field at a point above the sheet?
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Solution
The electric field due to an infinite plane sheet is constant and equal to E, regardless of the distance from the sheet.
Correct Answer: B — E
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Q. If the electric field due to a charged infinite plane sheet is E, what is the electric field at a point on either side of the sheet?
A.
E
B.
2E
C.
E/2
D.
Zero
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Solution
The electric field due to an infinite plane sheet is constant and equal to E on both sides, thus the total field is 2E.
Correct Answer: B — 2E
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Q. If the electric field due to a charged plane sheet is E, what is the electric field due to two parallel sheets with equal and opposite charge densities?
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Solution
The fields due to the two sheets add up in the region between them, resulting in a total electric field of 2E.
Correct Answer: C — 2E
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Q. If the electric field due to a point charge is 1000 N/C at a distance of 1m, what is the charge?
A.
1μC
B.
2μC
C.
3μC
D.
4μC
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Solution
Using E = k * |q| / r^2, we find q = E * r^2 / k = 1000 * 1^2 / (9 × 10^9) = 1μC.
Correct Answer: A — 1μC
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