Gauss Law
Q. 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 distance r from the center?
A.
0
B.
Q/4πε₀r²
C.
Q/4πε₀R²
D.
Q/4πε₀R
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Solution
For points outside the sphere, the electric field behaves as if all the charge were concentrated at the center, so E = Q/4πε₀r².
Correct Answer: B — Q/4πε₀r²
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Q. A cube encloses a charge Q at its center. What is the electric flux through one face of the cube?
A.
Q/ε₀
B.
Q/6ε₀
C.
Q/3ε₀
D.
Zero
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Solution
The total flux through the cube is Q/ε₀, and since there are 6 faces, the flux through one face is Q/6ε₀.
Correct Answer: B — Q/6ε₀
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Q. A cube of side length a has a charge Q at one of its corners. What is the electric flux through one face of the cube?
A.
Q/(6ε₀)
B.
Q/(12ε₀)
C.
Q/(8ε₀)
D.
Q/(4ε₀)
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Solution
The total flux through the cube is Q/ε₀, and since the charge is at one corner, the flux through one face is Q/(12ε₀).
Correct Answer: B — Q/(12ε₀)
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Q. A cube of side length a has a charge Q at one of its corners. What is the total electric flux through the cube?
A.
Q/ε₀
B.
Q/(6ε₀)
C.
Q/(12ε₀)
D.
0
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Solution
The charge at one corner contributes 1/8 of its flux to the cube. Thus, total flux = (1/8)Q/ε₀ = Q/(12ε₀).
Correct Answer: C — Q/(12ε₀)
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Q. A cylindrical conductor of radius R carries a uniform charge per unit length λ. What is the electric field at a distance r from the axis of the cylinder (r > R)?
A.
0
B.
λ/(2πε₀r)
C.
λ/(2πε₀R)
D.
λ/(4πε₀r²)
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Solution
For a point outside the cylinder, the electric field is given by E = λ/(2πε₀r).
Correct Answer: B — λ/(2πε₀r)
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Q. A cylindrical Gaussian surface encloses a charge Q. If the height of the cylinder is doubled while keeping the radius constant, what happens to the electric flux through the curved surface?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
The electric flux through the curved surface is proportional to the charge enclosed, which remains constant, so the flux through the curved surface doubles if the height is doubled.
Correct Answer: A — It doubles
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Q. A cylindrical Gaussian surface encloses a charge Q. If the radius of the cylinder is r and its height is h, what is the electric flux through the curved surface?
A.
Q/ε₀
B.
Q/(2ε₀)
C.
Q/(4ε₀)
D.
0
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Solution
The electric flux through the curved surface of a cylinder is given by Φ = Q_enc/ε₀, where Q_enc = Q.
Correct Answer: A — Q/ε₀
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Q. A cylindrical Gaussian surface encloses a charge Q. If the radius of the cylinder is doubled, what happens to the electric field at the surface?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It becomes zero
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Solution
The electric field due to a uniformly charged infinite cylinder depends only on the charge per unit length, not on the radius.
Correct Answer: C — It remains the same
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Q. A cylindrical Gaussian surface encloses a long straight wire carrying a current. What is the electric field at a distance r from the wire?
A.
0
B.
I/(2πε₀r)
C.
λ/(2πε₀r)
D.
σ/(2ε₀)
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Solution
Gauss's law applies to electric fields, not magnetic fields. The electric field around a current-carrying wire is not defined by Gauss's law.
Correct Answer: A — 0
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Q. A cylindrical Gaussian surface encloses a long straight wire carrying a current. What is the electric field at a point outside the cylinder?
A.
Zero
B.
Directly proportional to the distance from the wire
C.
Inversely proportional to the distance from the wire
D.
Constant
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Solution
The electric field outside the cylindrical surface is directly proportional to the distance from the wire.
Correct Answer: B — Directly proportional to the distance from the wire
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Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q uniformly distributed along its length. What is the electric field at a distance R from the axis of the cylinder?
A.
Q/(2πε₀R)
B.
Q/(4πε₀R²)
C.
0
D.
Q/(ε₀L)
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Solution
Using Gauss's law, the electric field outside the cylinder is E = Q/(2πε₀R).
Correct Answer: A — Q/(2πε₀R)
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Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q. What is the electric field E at a distance R from the axis of the cylinder?
A.
Q/(2πε₀R)
B.
Q/(4πε₀R²)
C.
Q/(ε₀L)
D.
0
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Solution
Using Gauss's law, the electric field E at a distance R from the axis of a long charged cylinder is E = Q/(2πε₀L) for points outside the cylinder.
Correct Answer: A — Q/(2πε₀R)
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Q. A hollow cylinder with charge density λ is placed along the z-axis. What is the electric field at a point outside the cylinder?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
Zero
D.
λ/(ε₀r)
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Solution
For a hollow cylinder, the electric field at a point outside is E = λ/(2πε₀r) using Gauss's law.
Correct Answer: A — λ/(2πε₀r)
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Q. A hollow sphere has a charge +Q distributed uniformly on its surface. What is the electric field at a point inside the sphere?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere has a charge Q distributed uniformly over its surface. What is the electric field inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀)
D.
Q/(4πε₀R)
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Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry.
Correct Answer: B — 0
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Q. A hollow sphere with charge Q has a radius R. What is the electric field at a point inside the sphere?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
D.
Q/(2πε₀R²)
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Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry, as per Gauss's law.
Correct Answer: B — 0
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Q. A hollow spherical conductor carries a charge Q. What is the electric field inside the cavity?
A.
Q/(4πε₀r²)
B.
0
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
Inside the cavity of a hollow conductor, the electric field is zero due to the shielding effect of the conductor.
Correct Answer: B — 0
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Q. A long straight wire carries a uniform linear charge density λ. What is the electric field at a distance r from the wire?
A.
λ/(2πε₀r)
B.
λ/(4πε₀r²)
C.
λ/(2πε₀r²)
D.
0
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Solution
Using Gauss's law for a cylindrical surface around the wire, the electric field E at a distance r is E = λ/(2πε₀r).
Correct Answer: A — λ/(2πε₀r)
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Q. 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/(4πε₀R²)
D.
Q/(4πε₀R)
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Solution
The 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. 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
The total electric flux through the surface is given by Gauss's law as Φ = Q/ε₀, and for a point charge at the center, it results in 4πQ/ε₀.
Correct Answer: D — 4πQ/ε₀
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Q. A point charge of +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/2ε₀
D.
4πQ/ε₀
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Solution
According to Gauss's law, the total electric flux through a closed surface is Φ = Q_enc/ε₀. Here, Q_enc = Q, so Φ = Q/ε₀.
Correct Answer: D — 4πQ/ε₀
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Q. A point charge of +Q is placed at the center of a spherical shell of radius R with surface charge density σ. What is the electric field inside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
σ/ε₀
D.
Q/(4πε₀R)
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Solution
According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
Correct Answer: A — 0
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Q. A point charge Q is placed at the center of a cube. What is the electric flux through one face of the cube?
A.
Q/ε₀
B.
Q/6ε₀
C.
Q/4ε₀
D.
0
Show solution
Solution
The total flux through the cube is Q/ε₀. Since there are 6 faces, the flux through one face is Q/(6ε₀).
Correct Answer: B — Q/6ε₀
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Q. A point charge Q is placed at the center of a spherical Gaussian surface. What is the electric flux through the surface?
A.
0
B.
Q/ε₀
C.
Q/4πε₀
D.
Q²/ε₀
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Solution
According to Gauss's law, the electric flux Φ = Q/ε₀ when a point charge Q is at the center of a spherical surface.
Correct Answer: B — Q/ε₀
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Q. A spherical Gaussian surface of radius R encloses a charge Q. What is the electric field at a distance 2R from the center?
A.
Q/4πε₀R²
B.
Q/4πε₀(2R)²
C.
0
D.
Q/ε₀(2R)²
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Solution
The electric field outside a spherical charge distribution is given by E = Q/4πε₀r². At 2R, it becomes Q/4πε₀(2R)².
Correct Answer: B — Q/4πε₀(2R)²
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Q. A spherical shell of radius R carries a total charge Q. What is the electric field at a point outside the shell?
A.
0
B.
Q/(4πε₀R²)
C.
Q/(4πε₀R)
D.
Q/(4πε₀R³)
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Solution
For a spherical shell, the electric field outside the shell behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).
Correct Answer: B — Q/(4πε₀R²)
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Q. A spherical shell of radius R carries a uniform charge Q. What is the electric field inside the shell?
A.
Q/(4πε₀R²)
B.
0
C.
Q/(4πε₀R)
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: B — 0
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Q. A spherical shell of radius R carries a uniform surface charge density σ. What is the electric field inside the shell?
A.
0
B.
σ/ε₀
C.
σ/2ε₀
D.
σ/4ε₀
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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. A uniformly charged sphere of radius R has a total charge Q. What is the electric field at a point outside the sphere (r > R)?
A.
0
B.
Q/(4πε₀r²)
C.
Q/(4πε₀R²)
D.
Q/(4πε₀R)
Show solution
Solution
For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center, thus E = Q/(4πε₀r²).
Correct Answer: B — Q/(4πε₀r²)
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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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