AP Physics 2- 10.5 Electric Potential- Exam Style questions - FRQs- New Syllabus

(a)
A particle accelerated through a potential difference gains kinetic energy equal to

\( K = |q\Delta V| \)

For Particle \( 1 \),

\( K_1 = |-Q\Delta V| = Q|\Delta V| \)

For Particle \( 2 \),

\( K_2 = |(+2Q)\Delta V| = 2Q|\Delta V| \)

Therefore,

\( \dfrac{K_2}{K_1} = \dfrac{2Q|\Delta V|}{Q|\Delta V|} = \boxed{2} \)

The masses do not affect this ratio here because the energy gain from an electric potential difference depends on charge and potential difference only.

(b)(i)
For Particle \( 2 \),

\( K_2 = \dfrac{1}{2}mv^2 \)

and the mass of Particle \( 2 \) is \( \dfrac{M}{2} \), so

\( K_2 = \dfrac{1}{2}\left(\dfrac{M}{2}\right)v^2 = \dfrac{Mv^2}{4} \)

Solving for \( v \),

\( v^2 = \dfrac{4K_2}{M} \)

\( \boxed{v = 2\sqrt{\dfrac{K_2}{M}}} \)

(b)(ii)
In the magnetic field, the magnetic force provides the centripetal force:

\( qvB_0 = \dfrac{mv^2}{r} \)

so

\( r = \dfrac{mv}{qB_0} \)

For Particle \( 2 \),

\( m = \dfrac{M}{2}, \qquad q = 2Q, \qquad v = 2\sqrt{\dfrac{K_2}{M}} \)

Therefore,

\( r = \dfrac{\left(\dfrac{M}{2}\right)\left(2\sqrt{\dfrac{K_2}{M}}\right)}{2QB_0} \)

\( r = \dfrac{\sqrt{MK_2}}{2QB_0} \)

Since the particle enters moving downward and exits moving upward, it travels through a semicircle, so the horizontal distance between entry and exit points is the diameter:

\( \Delta x = 2r \)

Thus,

\( \boxed{\Delta x = \dfrac{\sqrt{MK_2}}{QB_0}} \)

(c)
Using \( \vec{F}_B = q\vec{v}\times\vec{B} \):

The particles enter with velocity in the \( -y \)-direction and the magnetic field is in the \( +z \)-direction.

For a positive charge, \( \vec{v}\times\vec{B} \) points in the \( -x \)-direction, so Particle \( 2 \) curves to the left.

For a negative charge, the force is opposite that direction, so Particle \( 1 \) curves to the right.

Also, Particle \( 1 \) has the larger radius of curvature. One quick reason is:

\( r=\dfrac{mv}{|q|B} \)

and here Particle \( 1 \) has a smaller charge magnitude than Particle \( 2 \), so it bends less.

(d)
For Particle \( 1 \) to move in a straight line at constant speed, the net force must be zero. Therefore, the electric force must cancel the magnetic force.

For Particle \( 1 \), the magnetic force is toward the \( +x \)-direction. So the electric force must be toward the \( -x \)-direction.

Since Particle \( 1 \) has charge \( -Q \), its electric force is opposite the electric field direction. Therefore, for the electric force to be toward \( -x \), the electric field must point toward \( +x \).

\( \boxed{\text{Electric field direction: } +x\text{-direction}} \)