Q. A charge of +3μC is placed in a uniform electric field of strength 1500 N/C. What is the work done in moving the charge 0.2 m in the direction of the field?
A.90 J
B.60 J
C.30 J
D.45 J
Solution
Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.
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
Solution
For points outside the sphere, the electric field behaves as if all the charge were concentrated at the center, so E = Q/4πε₀r².
Q. A charged particle moves from a point of higher electric potential to a point of lower electric potential. What happens to its kinetic energy?
A.Increases
B.Decreases
C.Remains constant
D.Cannot be determined
Solution
As the charged particle moves to a lower potential, it loses potential energy, which is converted into kinetic energy, thus increasing its kinetic energy.
Q. A charged particle moves from a region of high potential to low potential. What happens to its kinetic energy?
A.It increases
B.It decreases
C.It remains constant
D.It becomes zero
Solution
As the charged particle moves from high potential to low potential, it loses potential energy, which is converted into kinetic energy, thus its kinetic energy increases.
Q. A charged particle moves in a magnetic field B with a velocity v. What is the expression for the magnetic force acting on the particle?
A.qvB
B.qvBsinθ
C.qvBcosθ
D.qB
Solution
The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvBsinθ, where θ is the angle between the velocity and the magnetic field.
Q. A charged particle moves in a magnetic field. What is the condition for the particle to experience maximum force?
A.Velocity is zero
B.Velocity is parallel to the field
C.Velocity is perpendicular to the field
D.Charge is zero
Solution
The magnetic force on a charged particle is given by F = qvB sin(θ). The force is maximum when the angle θ is 90 degrees, meaning the velocity is perpendicular to the magnetic field.
Correct Answer: C — Velocity is perpendicular to the field
Q. A charged particle moves in a magnetic field. What is the condition for the particle to experience no magnetic force?
A.The particle must be at rest
B.The particle must be moving parallel to the magnetic field
C.The particle must be moving perpendicular to the magnetic field
D.The magnetic field must be zero
Solution
The magnetic force on a charged particle is given by F = q(v × B). If the velocity vector v is parallel to the magnetic field B, the cross product is zero, resulting in no magnetic force.
Correct Answer: B — The particle must be moving parallel to the magnetic field
Q. A charged particle moves in a magnetic field. What is the effect of the magnetic field on the particle's motion?
A.It accelerates the particle
B.It changes the particle's speed
C.It changes the particle's direction
D.It has no effect
Solution
A magnetic field exerts a force on a charged particle that is perpendicular to both the velocity of the particle and the magnetic field, changing its direction but not its speed.
Correct Answer: C — It changes the particle's direction
Q. A charged particle moves in a magnetic field. What is the nature of the force acting on it?
A.Always in the direction of motion
B.Always opposite to the direction of motion
C.Perpendicular to the direction of motion
D.Depends on the charge of the particle
Solution
The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force law, which states that the force is perpendicular to both the velocity of the particle and the magnetic field.
Correct Answer: C — Perpendicular to the direction of motion
Q. A charged sphere has a radius R and a total charge Q. What is the electric potential at a point outside the sphere at a distance r from the center (r > R)?
A.kQ/R
B.kQ/r
C.kQ/(R+r)
D.0
Solution
For a charged sphere, the electric potential outside the sphere behaves as if all the charge were concentrated at the center, so V = kQ/r.
