AP Physics 2- 10.4 Electric Potential Energy- Exam Style questions - FRQs- New Syllabus

(a)
Since sphere \(Y\) is positive and the outermost equipotential line has magnitude \(50\,\text{V}\), the potentials shown are positive.

Adjacent equipotential lines differ by \(10\,\text{V}\). Point \(A\) is on the \(60\,\text{V}\) line, and point \(B\) is on the \(80\,\text{V}\) line.

\(\boxed{V_A=+60\,\text{V}}\)

\(\boxed{V_B=+80\,\text{V}}\)

(b)(i)
Sphere \(Y\) has a greater magnitude of charge than sphere \(X\).

The equipotential lines are more spread out and extend farther around sphere \(Y\), showing that \(Y\) produces a larger electric potential contribution. Since the charge on \(Y\) is positive and there is no zero-potential line between the spheres, sphere \(X\) must also be positive. If the charges had opposite signs, there would be a location where the electric potentials cancel to zero.

\(\boxed{\text{Both spheres are positive, and } |q_Y|>|q_X|.}\)

(b)(ii)
The gravitational potential isolines have similar shapes because gravitational potential and electric potential have similar distance dependence.

Electric potential from a point charge depends on \(\dfrac{q}{r}\), while gravitational potential depends on \(\dfrac{m}{r}\). Since the masses are in the same ratio as the magnitudes of the charges, the relative strengths of the two sources are similar, so the isoline patterns have similar shapes.

(c)
A similarity is that both the electric force and gravitational force depend on distance in the same inverse-square way, proportional to \(\dfrac{1}{r^2}\). Therefore, both forces are stronger when the proton is closer to a sphere.

A difference is that the proton and the positively charged spheres electrically repel each other, but gravitational forces are attractive. Therefore, the electric force on the proton points generally away from the positive spheres, while the gravitational force points toward the spheres.

(d)(i)
The proton is positive, so it is repelled by the positive spheres and moves toward lower electric potential. Since it starts at \(B\) and moves through a potential difference of \(20\,\text{V}\), it moves from \(V_B=80\,\text{V}\) to \(V_A=60\,\text{V}\).

The change in electric potential energy is

\(\Delta U_E=q\Delta V\)

\(\Delta U_E=q\left(V_A-V_B\right)\)

Since \(V_A-V_B=-20\,\text{V}\),

\(\boxed{\Delta U_E=q\left(V_A-V_B\right)=-20q}\)

(d)(ii)
Work done by the electric field is

\(W=Fd\)

The electric force on the proton has average magnitude

\(F_{\text{avg}}=qE_{\text{avg}}\)

Therefore,

\(\boxed{W=qE_{\text{avg}}d}\)

This is also consistent with \(W=-\Delta U_E\). Since \(\Delta U_E=-20q\), the electric field does positive work \(W=20q\).

(d)(iii)
Student \(1\) is correct. If the system is defined as the proton and the spheres, then the electric interaction is internal to the system. The proton’s kinetic energy increases because electric potential energy of the system decreases. Energy is transformed from electric potential energy into kinetic energy.

Student \(2\) is also correct. If the system is defined as only the proton, then the electric field from the spheres is external to the system. The electric field does positive work on the proton as it moves through the \(20\,\text{V}\) potential difference, and this positive work increases the proton’s kinetic energy.

\(\boxed{\text{Both claims are correct, but they use different system definitions.}}\)