Gauss Law
Q. For a charged plane sheet, if the surface charge density is doubled, what happens to the electric field?
A.
It remains the same
B.
It doubles
C.
It halves
D.
It quadruples
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Solution
The electric field due to a charged plane sheet is directly proportional to the surface charge density. Therefore, if σ is doubled, E also doubles.
Correct Answer: B — It doubles
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Q. For a charged sphere, what happens to the electric field inside the sphere as the radius increases?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
The electric field inside a uniformly charged sphere is zero.
Correct Answer: D — Becomes zero
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Q. For a charged spherical conductor, what happens to the electric field inside the conductor when it is charged?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
The electric field inside a charged conductor in electrostatic equilibrium is zero.
Correct Answer: D — Becomes zero
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Q. For a closed surface enclosing multiple charges, how is the total electric flux calculated?
A.
Sum of individual fluxes
B.
Product of charges
C.
Sum of enclosed charges divided by ε₀
D.
Average of charges
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Solution
The total electric flux through a closed surface is given by Φ = ΣQ_enc/ε₀, where Q_enc is the total charge enclosed.
Correct Answer: C — Sum of enclosed charges divided by ε₀
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Q. For a closed surface enclosing multiple charges, how is the total electric flux related to the enclosed charges?
A.
It is proportional to the sum of the charges
B.
It is inversely proportional to the sum of the charges
C.
It is independent of the charges
D.
It is proportional to the square of the charges
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Solution
According to Gauss's law, the total electric flux through a closed surface is proportional to the total charge enclosed.
Correct Answer: A — It is proportional to the sum of the charges
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Q. For a point charge, the electric field varies with distance r as?
A.
1/r
B.
1/r²
C.
1/r³
D.
1/r⁴
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Solution
The electric field due to a point charge varies as E = kQ/r², where k is a constant.
Correct Answer: B — 1/r²
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Q. For a spherical Gaussian surface of radius R enclosing a charge Q, what is the electric field at a distance 2R from the center?
A.
Q/4πε₀(2R)²
B.
Q/4πε₀R²
C.
Q/4πε₀(2R)³
D.
0
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Solution
The electric field outside a spherical charge distribution behaves as if all the charge were concentrated at the center, so E = Q/4πε₀r².
Correct Answer: A — Q/4πε₀(2R)²
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Q. For a uniformly charged sphere of radius R and total charge Q, what is the electric field at a distance r from the center where r > R?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
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, given by E = Q/(4πε₀r²).
Correct Answer: A — Q/(4πε₀r²)
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Q. For an infinite plane sheet of charge with surface charge density σ, what is the electric field at any point?
A.
σ/2ε₀
B.
σ/ε₀
C.
0
D.
σ/4πε₀
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Solution
The electric field due to an infinite plane sheet of charge is constant and given by E = σ/2ε₀ on either side of the sheet.
Correct Answer: A — σ/2ε₀
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Q. For an infinite plane sheet of charge with surface charge density σ, what is the electric field at a point near the sheet?
A.
σ/2ε₀
B.
σ/ε₀
C.
0
D.
σ/4πε₀
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Solution
Using Gauss's law, the electric field due to an infinite plane sheet of charge is E = σ/2ε₀ on either side of the sheet.
Correct Answer: A — σ/2ε₀
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Q. If a charge of +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
The total flux through the cube is Q/ε₀. Since the charge is at one corner, the flux through one face is Q/6ε₀.
Correct Answer: A — Q/6ε₀
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Q. If a charge of +Q is placed at one corner of a cube, what is the total electric flux through the entire surface of the cube?
A.
Q/ε₀
B.
Q/6ε₀
C.
0
D.
Q/4ε₀
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Solution
The total electric flux through the cube is Q/ε₀, but since only 1/6 of the charge is enclosed by the cube, the flux is Q/6ε₀.
Correct Answer: B — Q/6ε₀
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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 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 E, what is the electric field at a distance of 2r from the charge?
A.
E/2
B.
E/4
C.
E/8
D.
E
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Solution
The electric field due to a point charge varies inversely with the square of the distance, so at 2r, E becomes E/4.
Correct Answer: B — E/4
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Q. If the electric field inside a conductor in electrostatic equilibrium is zero, what can be said about the charge distribution?
A.
Charge is uniformly distributed
B.
Charge is concentrated at the center
C.
Charge resides on the surface
D.
Charge is absent
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Solution
In electrostatic equilibrium, excess charge resides on the surface of the conductor, leading to zero electric field inside.
Correct Answer: C — Charge resides on the surface
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