Electrostatics
Q. A dipole consists of two charges +q and -q separated by a distance d. What is the expression for the dipole moment?
A.
qd
B.
q/d
C.
q^2d
D.
q/d^2
Show solution
Solution
The dipole moment p is defined as p = q * d.
Correct Answer: A — qd
Learn More →
Q. A dipole consists of two charges +q and -q separated by a distance d. What is the dipole moment?
A.
qd
B.
q/d
C.
q^2d
D.
q/d^2
Show solution
Solution
The dipole moment p = q * d.
Correct Answer: A — qd
Learn More →
Q. A dipole consists of two equal and opposite charges separated by a distance of 0.1m. What is the dipole moment if each charge is 1μC?
A.
1 × 10^-7 C m
B.
1 × 10^-6 C m
C.
1 × 10^-5 C m
D.
1 × 10^-4 C m
Show solution
Solution
Dipole moment p = q * d = (1 × 10^-6 C) * (0.1 m) = 1 × 10^-7 C m.
Correct Answer: B — 1 × 10^-6 C m
Learn More →
Q. A dipole consists of two equal and opposite charges separated by a distance. What happens to the dipole moment if the distance is doubled?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
Show solution
Solution
Dipole moment p = q * d, if d is doubled, p also doubles.
Correct Answer: A — It doubles
Learn More →
Q. A dipole moment is defined as the product of charge and the distance between the charges. What is the dipole moment of a dipole consisting of charges +2μC and -2μC separated by 0.1m?
A.
4 × 10^-7 C m
B.
2 × 10^-7 C m
C.
2 × 10^-6 C m
D.
4 × 10^-6 C m
Show solution
Solution
Dipole moment p = q * d = (2 × 10^-6 C) * (0.1 m) = 2 × 10^-7 C m.
Correct Answer: A — 4 × 10^-7 C m
Learn More →
Q. A dipole moment p is placed in a uniform electric field E. What is the torque experienced by the dipole?
A.
pE
B.
pE sin θ
C.
pE cos θ
D.
0
Show solution
Solution
Torque τ = p × E = pE sin θ, where θ is the angle between p and E.
Correct Answer: B — pE sin θ
Learn More →
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)
Show solution
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)
Learn More →
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)
Show solution
Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
Learn More →
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)
Show solution
Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry.
Correct Answer: B — 0
Learn More →
Q. A hollow sphere has a charge Q uniformly distributed on its surface. What is the electric field inside the sphere?
A.
Q/4πε₀R²
B.
0
C.
Q/ε₀
D.
Q/4πε₀
Show solution
Solution
The electric field inside a hollow charged sphere is zero due to symmetry.
Correct Answer: B — 0
Learn More →
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)
Show solution
Solution
Inside a hollow charged sphere, the electric field is zero due to symmetry, as per Gauss's law.
Correct Answer: B — 0
Learn More →
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)
Show solution
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
Learn More →
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
Show solution
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)
Learn More →
Q. A parallel plate capacitor has a potential difference of V across its plates. What is the electric field between the plates?
A.
V/d
B.
d/V
C.
V²/d
D.
d²/V
Show solution
Solution
The electric field E between the plates of a parallel plate capacitor is given by E = V/d, where d is the separation between the plates.
Correct Answer: A — V/d
Learn More →
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)
Show solution
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
Learn More →
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
Show solution
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
Learn More →
Q. A point charge of +5 µC is placed at the origin. What is the electric potential at a point 2 m away from the charge?
A.
1125 V
B.
450 V
C.
225 V
D.
0 V
Show solution
Solution
The electric potential V at a distance r from a point charge Q is given by V = kQ/r. Here, V = (9 x 10^9 Nm²/C²)(5 x 10^-6 C)/(2 m) = 1125 V.
Correct Answer: A — 1125 V
Learn More →
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
Show solution
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
Learn More →
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)
Show solution
Solution
According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
Correct Answer: A — 0
Learn More →
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
Learn More →
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
Show solution
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
Learn More →
Q. A spherical conductor has a charge Q. What is the electric potential inside the conductor?
A.
0
B.
Q/(4πε₀r)
C.
Q/(4πε₀R)
D.
Constant throughout
Show solution
Solution
The electric potential inside a charged spherical conductor is constant and equal to the potential on its surface, which is Q/(4πε₀R).
Correct Answer: D — Constant throughout
Learn More →
Q. A spherical conductor has a radius R and carries a charge Q. What is the electric potential on its surface?
A.
kQ/R
B.
kQ/2R
C.
0
D.
kQ/R²
Show solution
Solution
The electric potential V on the surface of a charged spherical conductor is given by V = kQ/R.
Correct Answer: A — kQ/R
Learn More →
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
Show solution
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
Learn More →
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)
Show solution
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)
Learn More →
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)
Show solution
Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer: B — 0
Learn More →
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
Show solution
Solution
By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.
Correct Answer: A — 0
Learn More →
Q. A uniform electric field of 200 N/C is present. What is the potential difference between two points 3 m apart?
A.
600 V
B.
400 V
C.
200 V
D.
800 V
Show solution
Solution
The potential difference V = E * d = 200 N/C * 3 m = 600 V.
Correct Answer: A — 600 V
Learn More →
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)
Learn More →
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
Show solution
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
Learn More →
Showing 61 to 90 of 363 (13 Pages)