Home / AP Physics C Electricity and Magnetism – 1.1 Electrostatics: Charge  and Coulomb’s Law -FRQ Exam style question

AP Physics C Electricity and Magnetism – 1.1 Electrostatics: Charge  and Coulomb’s Law -FRQ Exam style question

Question 

1. A nonconducting rod of uniform positive linear charge density is near a sphere with charge −2.0 nC. The rod and sphere are held at rest on the x-axis, as shown in Figure 1. Equipotential lines and positions A, B, C, D, and E are labeled. Adjacent tick marks on the x-axis and the y-axis are 0.40 m apart.

(a) Calculate the absolute value of the electric flux through the Gaussian surface whose cross section is the −20.0 V equipotential line.

A positive test charge (not shown) is placed and held at rest at Position C. An external force is applied to the test charge to move the test charge to different positions in the order of C→ →E D→A. The test charge is momentarily held at rest at each position.

(b) The bar shown in Figure 2 represents the absolute value of the work \( W_{CE}\) done by the external force on the test charge to move the test charge from Position C to Position E.

  i. Complete the following tasks on Figure 2.
  • Draw a bar to represent the relative absolute value of the work \( W_{CD}\) done by the external force on the test charge to move the test charge from Position E to Position D.
 • Draw a bar to represent the relative absolute value of the work \( W_{DA}\) done by the external force on the test charge to move the test charge from Position D to Position A.
 • The height of each bar should be proportional to the value of \( W_{CE}\). If  \( W_{ED}\) = and/or \( W_{DA}\) = 0, write a “0” in the corresponding columns, as appropriate.

ii. Calculate the approximate magnitude of the x-component of the electric field at Position B.

The positive test charge is placed at Position D. The test charge is then released from rest.
(c) Indicate the direction (not components) of the net electric force exerted on the test charge immediately after the test charge is released from rest.

_____ +x _____ +y _____ Directly away from the sphere
_____ −x _____ −y _____ Directly toward the sphere

Without using equations, justify your answer using physics principles.

The sphere and the test charge are removed. The rod has length 4L and uniform positive linear charge density \(+\lambda \). The rod is held at rest on the x-axis in the orientation shown in Figure 3. Position P (not shown) is located on the x-axis a distance \(x_{p}\) from the origin, where \(x_{p}\) > 4L.

(d) The electric potential \(V_{p}\) at \(x_{p}\) is \(V_{p}\) = \(K\lambda \) In\(\left ( \frac{x_{p}}{x_{p} – 4L} \right )\).

i. Using integral calculus, derive the expression for \(V_{p}\) provided.

ii. On Figure 4, sketch a graph of the x-component \(E_{x}\) of the electric field from the rod as a function of x in the region 4L < x < 12L.

▶️Answer/Explanation

1(a) Example Solution

1(b) (i) Example Response

1(b) (ii) Example Solution

1(c) Example Response 

+y. The direction of an electric field vector is perpendicular to an equipotential line. Because the test charge has a positive charge, the test charge would move from a position of higher electric potential to a position of lower electric potential when an electric force is exerted on the test charge. Therefore, at Position D, the electric force is upward because that is the direction that is perpendicular to the equipotential line and in the direction of decreasing electric potential.

1(d) (i) Example Solutions

OR

1(d) (ii) Example Response

Question 

1. A nonconducting rod of uniform positive linear charge density is near a sphere with charge −2.0 nC. The rod and sphere are held at rest on the x-axis, as shown in Figure 1. Equipotential lines and positions A, B, C, D, and E are labeled. Adjacent tick marks on the x-axis and the y-axis are 0.40 m apart.

(a) Calculate the absolute value of the electric flux through the Gaussian surface whose cross section is the −20.0 V equipotential line.

A positive test charge (not shown) is placed and held at rest at Position C. An external force is applied to the test charge to move the test charge to different positions in the order of C→ →E D→A. The test charge is momentarily held at rest at each position.

(b) The bar shown in Figure 2 represents the absolute value of the work \( W_{CE}\) done by the external force on the test charge to move the test charge from Position C to Position E.

  i. Complete the following tasks on Figure 2.
  • Draw a bar to represent the relative absolute value of the work \( W_{CD}\) done by the external force on the test charge to move the test charge from Position E to Position D.
 • Draw a bar to represent the relative absolute value of the work \( W_{DA}\) done by the external force on the test charge to move the test charge from Position D to Position A.
 • The height of each bar should be proportional to the value of \( W_{CE}\). If  \( W_{ED}\) = and/or \( W_{DA}\) = 0, write a “0” in the corresponding columns, as appropriate.

ii. Calculate the approximate magnitude of the x-component of the electric field at Position B.

The positive test charge is placed at Position D. The test charge is then released from rest.
(c) Indicate the direction (not components) of the net electric force exerted on the test charge immediately after the test charge is released from rest.

_____ +x _____ +y _____ Directly away from the sphere
_____ −x _____ −y _____ Directly toward the sphere

Without using equations, justify your answer using physics principles.

The sphere and the test charge are removed. The rod has length 4L and uniform positive linear charge density \(+\lambda \). The rod is held at rest on the x-axis in the orientation shown in Figure 3. Position P (not shown) is located on the x-axis a distance \(x_{p}\) from the origin, where \(x_{p}\) > 4L.

(d) The electric potential \(V_{p}\) at \(x_{p}\) is \(V_{p}\) = \(K\lambda \) In\(\left ( \frac{x_{p}}{x_{p} – 4L} \right )\).

i. Using integral calculus, derive the expression for \(V_{p}\) provided.

ii. On Figure 4, sketch a graph of the x-component \(E_{x}\) of the electric field from the rod as a function of x in the region 4L < x < 12L.

▶️Answer/Explanation

1(a) Example Solution

1(b) (i) Example Response

1(b) (ii) Example Solution

1(c) Example Response 

+y. The direction of an electric field vector is perpendicular to an equipotential line. Because the test charge has a positive charge, the test charge would move from a position of higher electric potential to a position of lower electric potential when an electric force is exerted on the test charge. Therefore, at Position D, the electric force is upward because that is the direction that is perpendicular to the equipotential line and in the direction of decreasing electric potential.

1(d) (i) Example Solutions

OR

1(d) (ii) Example Response

 Question

Four charge \(+\mathrm{q},+\mathrm{q},-\mathrm{q}\) and \(-\mathrm{q}\) are placed respectively at the corners \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and D of a square of side (a), arranged in the given order. Calculate the intensity of electric field at the center of the square where \(Q\) is a charge.

▶️Answer/Explanation

Ans:

Consider the problem


\(\mathrm{F}_1\) by \(-\mathrm{q}\) and \(+\mathrm{q}\) on \(\mathrm{Q}\) charge
\[
\begin{aligned}
& \mathrm{F}_1=\frac{\mathrm{kQq}}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)^2}+\frac{\mathrm{kQq}}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)^2} \\
& =\frac{4 \mathrm{kQq}}{\mathrm{a}^2} \mathrm{~N}
\end{aligned}
\]

And
\(F_2\) by \(-q\) and \(+q\) on \(Q\) charge
\[
\begin{aligned}
& F_2=\frac{k Q q}{\left(\frac{a}{\sqrt{2}}\right)^2}+\frac{k Q q}{\left(\frac{a}{\sqrt{2}}\right)^2} \\
& =\frac{4 k Q q}{a^2} \mathrm{~N} \\
& F_1=F_2=\frac{4 k Q q}{a^2} N
\end{aligned}
\]

Now F Resultant
\[
\begin{aligned}
& =\sqrt{\left(F_1\right)^2+\left(F_2\right)^2+2 F_1 F_2 \cos 90^{\circ}} \\
& =\sqrt{2} \mathrm{~F} \\
& =\sqrt{2}\left(\frac{4 \mathrm{kQq}}{\mathrm{a}^2}\right) \\
& =4 \sqrt{2} \frac{\mathrm{kQq}}{\mathrm{a}^2}
\end{aligned}
\]

 

 Question

Students perform an experiment to study the force between two charged objects using the apparatus shown above, which contains two identical conducting spheres. The upper sphere is attached to an insulating string, which can be used to move the sphere downward. The lower sphere sits on an insulating rod, which is on an electronic balance. The electronic balance is zeroed before the lower sphere and insulating rod are in place. For the first trial, a charge of Q is placed on each sphere and then the upper sphere is slowly moved downward. The students measure the distance d between the centers of the spheres and the magnitude F of the force that appears on the electronic balance. The recorded data are shown on the graph of F as a function of \(\frac{1}{d^{2}}\)  shown below.

(a)
i. Draw a line that represents the best fit to the points shown.
ii. Use the graph to calculate the charge Q.
iii. On the graph on the previous page, draw a circle around the data point that was taken when the distance between the centers of the spheres was the least.
iv. Determine the distance between the centers of the spheres for the data point indicated above.
v. What physical quantity does the vertical intercept represent?
Justify your answer.

The experiment is extended by collecting additional data points, which appear on the right side of the graph shown above. The new data points do not follow the linear pattern seen with the first points. The group of students tries to explain this discrepancy.
(b) One student suspects that charge is slowly leaking off the top sphere. Could this explain the discrepancy?
____ Yes _____ No
Justify your answer.

(c) A second student suspects that the excess charges have rearranged themselves, polarizing the spheres.
i. On the circles representing the spheres below, use a single “ ”+ sign on each sphere to represent the locations of highest concentration of the excess positive charges.

ii. Explain how this rearrangement could be responsible for the discrepancy.
(d) A third student suggests that the experiment be modified so that the top sphere is given a negative charge that is equal in magnitude to the positive charge given to the bottom sphere.
i. On the circles representing the spheres below, use a single “ ”+ sign on the bottom sphere to represent the location of highest concentration of the excess positive charges. Use a single “ ” − sign on the top sphere to represent the location of the highest concentration of the excess negative charges.

ii. For a separation distance equal to that of the data point indicated in part (a)(iii), would the magnitude of the force reading with spheres of opposite charges be greater than, less than, or equal to the magnitude of the force reading with spheres of the same charges?
_____ Greater than _____ Less than _____ Equal to
Justify your answer.

Answer/Explanation

Ans:

(a) ii. 

\(slope = \frac{100.500\mu N}{60\frac{1}{m^{2}}}=0.69 Nm^{2}\)

\(Q = \sqrt{\frac{Fd^{2}}{k}}=\sqrt{\frac{slope}{k}}=8.76 \mu C\)

iv. 

\(\frac{1}{d^{2}}= 81\frac{1}{m^{2}}\)                        d = 0.111 m

v.

the weight of the lower sphere and insulating rod 

The vertical intercept represents the force when the top sphere is infinitely for away, so the only force acting on the scale is the weight of the lower sphere and insulating rod.

(b)

   X    Yes

If the top sphere’s charge decreases, then the force also decreases, as seen in the graph because of the curve of best fit falling below a straight line. Therefore, leaking charge could explain the discrepancy. 

(c) i. 

ii.

   X    Less than

With opposite charges, the electric force and weight force oppose each other, which decreases the magnitude of the force when compared to the same charges.

Question

A rod of uniform linear charge density \(\lambda =+1.5 \times 10^{-5}\) C/m is bent  into an arc of radius R = 0.10 m. The arc is placed with its center at the origin of the axes shown above.
a. Determine the total charge on rod.
b. Determine the magnitude and direction of the electric field at the center O of the arc.
c. Determine the electric potential at point O.
A proton is now placed at point O and held in place. Ignore the effects of gravity in the rest of this problem.
d. Determine the magnitude and direction of the force that must be applied in order to keep the proton at
rest.
e. The proton is now released. Describe in words its motion for a long time after its release.

Answer/Explanation

Ans:

a. The total charge is \(q=\lambda L = \lambda (2 \piR/3)=3.1 \times 10^{-6}\) C
b. \(E=kQ/R^2\), all of the charge is equidistant, but we must take direction into account. The y components will
cancel so we only need to consider the x (horizontal) component from each infinitesimal element dq = λR dθ
\(E=\int_{120}^{240}\frac{kdq}{R^2}cos \theta d\theta = \int_{120}^{240}\frac{kd \lambda R}{R^2}cos \theta d \theta =\int_{120}^{240}\frac{k\lambda}{R}cos \theta d\theta=\frac{k\lambda}{R}sin \theta |^{240}_{120}=2.3 \times 10^6 N/C\)
To the right
c. \(V=kQ/R=2.8 \times 10^5\) V ( all charge is equidistant from point O)
d. \(F=qE=3.7 \times 10^{-13}N\)
e. The proton moves off to the right, but as the force decreases the proton’s acceleration decreases, all the while
speeding up to the right asymptotic to some value

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