AP Physics 2: 3.13 Conservation of Electric Energy – Exam Style questions with Answer- MCQ

Question

The nonconducting hollow sphere of radius R shown above carries a large charge +Q, which is uniformly distributed on its surface. There is a small hole in the sphere. A small charge +q is initially located at point P. a distance r from the center of the sphere. If k = 1/4πεo, what is the work that must be done by an external agent in moving the charge +q from P through the hole to the center O of the sphere?
(A) kqQ/r               (B) kqQ/R                   (C) kq(Q – q)/r                      (D) kqQ(1/R – 1/r)

▶️Answer/Explanation

Ans:D

Solution: Inside the sphere, E = 0 which means the potential does not change with position and is the same value as the surface, which is kQ/R. At point P, the potential is kQ/r. W = qΔV = q(kQ/R – kQ/r)

Question

A positive charge of 3.0 × 10–8 coulomb is placed in an upward directed uniform electric field of 4.0 × 104 N/C. When the charge is moved 0.5 meter upward, the work done by the electric force on the charge is
(A) 6 × 10–4 J                (B) 12 × 10–4 J                   (C) 2 × 104 J                    (D) 8 × 104 J

▶️Answer/Explanation

Ans:A

Solution: W = Fd = qEd

Question

Points R and S are each the same distance d from two unequal charges, +Q and +2Q, as shown above. The work
required to move a charge –Q from point R to point S is
(A) dependent on the path taken from R to S
(B) positive
(C) zero
(D) negative

▶️Answer/Explanation

Ans:C

Solution: By symmetry VR = VS so ΔVRS = 0 and W = qΔV

Question

A particle of charge Q and mass m is accelerated from rest through a potential difference V, attaining a kinetic energy K. What is the kinetic energy of a particle of charge 2Q and mass m/2 that is accelerated from rest through the same potential difference?
(A) ½ K               (B) K                (C) 2K                   (D) 4K

▶️Answer/Explanation

Ans:C

Solution: W = ΔK = QV (mass doesn’t have an effect on the kinetic energy, just on the speed in this case)

Question

A helium nucleus (charge +2q and mass 4m) and a lithium nucleus (charge +3q and mass 7m) are accelerated through the same electric potential difference, V0. What is the ratio of their resultant kinetic energies, \(\frac{K_{lithium}}{K_{Helium}}\)  ?
(A) 2/3             (B) 6/7              (C) 7/6                (D) 3/2

▶️Answer/Explanation

Ans:D

Solution: K = qΕV so K1/K2 = q1/q2

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