AP® Physics C: E&M - Electric Flux - FRQs
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
A thin wire of length L has a uniform charge density \(=\lambda \). A cylindrical Gaussian surface of radius d is drawn with the wire along its central axis, as shown above. Point P is located at the center of one end of the cylinder, a distance d from the end of the wire. Point Q is on the edge of the cylinder directly above the center of the wire, as shown above. A student says, “Gauss’s law can be used to find the electric flux \( \Phi \)through the Gaussian surface.”
(a) Is the student’s statement correct or incorrect? ____ Correct ____ Incorrect If you have chosen “Correct,” use Gauss’s law to find the electric flux F through the Gaussian surface. If you have chosen “Incorrect,” explain why the student’s reasoning is incorrect and why Gauss’s law cannot be applied in this situation.
(b) Two students discuss whether or not they can use Gauss’s law to find the electric field at points P and Q. At which of the points, if either, is Gauss’s law a useful method for finding the electric field? _____ At point P only ____ At point Q only _____ At both points P and Q ____ At neither point P nor point Q Justify your answer.
(c) Assuming the electric potential is zero at infinity, show that the value for the electric potential at point P is given by the following expression.
\(v=\frac{\lambda }{4\pi \varepsilon _0}In(\frac{L+d}{d})\)
The wire is aligned along the x-axis with the origin at the left end of the wire, as shown in Figure 2 above.
(d) A positively charged particle of charge +e and mass m is released from rest at point P. On the axes below, sketch the kinetic energy K of the particle, the potential energy U of the wire-particle system, and the total energy \(E_{tot}\) of the wire-particle system as functions of the particle’s position x. Clearly label each sketch with K, U, and E tot . Explicitly label any maximum with numerical values or algebraic expressions, as appropriate.
(e) Derive an expression for the magnitude of the electric field due to the wire as a function of the position along the x-axis, where x > L . Express your answer in terms of x, L, l , and physical constants, as appropriate.
Answer/Explanation
(a) Select “Correct” Note: If the wrong selection is made, the explanation is ignored. For using an appropriate equation to calculate the flux
Claim: Student is correct.
Evidence: A cylinder is useful for Gauss’s law.
Reasoning: A cylindrical surface has geometric symmetry.
(b) Select “At neither point P nor point Q” or “At point Q only”
Note: If the wrong selection is made, the justification is ignored.
For a justification consistent with selection above
Example: There is no simple way to write the electric field at point P or Q in terms of
the flux due to cylinder extending beyond the line of charge.
Example: By drawing a new Gaussian cylinder that does not extend beyond the line of
charge, Gauss’s law can be used to calculate the electric field at point Q.
(c) For indicating an attempt to integrate to determine the electric potential at P\( V=\int \frac{1}{4\pi\varepsilon _0r}dq\)
For integrating in terms of distance
\(Q=\lambda L\therefore =dq=\lambda dr\therefore V=\int_{r=d+L}^{r=d}\frac{\lambda }{4\pi\varepsilon _0}dr\)
For integrating using the correct limits or constant of integration \(V=\frac{\lambda }{4\pi\varepsilon _0}[ln(r)]_{r=d}^{r=d+L}=\frac{\lambda }{4\pi\varepsilon _0}(ln(d+L))-ln(d))=\frac{\lambda }{4\pi\varepsilon _0}ln\left ( \frac{d+L}{d} \right )\)
(d) For a curve in the first quadrant label K that is increasing in value for x >L + d point For a curve that is concave down curve starting at the point ( L +d , 0) and approaching a horizontal line
For a curve in the first quadrant label U that is decreasing in value for x> L + d point For a concave up curve at L + d starting at or near the maximum value and approaching the x-axis
For a horizontal line labeled\( E_{tot}\) that is at the maximum value 1 point For labeling and using correct asymptotes for the K and U curves
(e)
Question
A nonconducting sphere with center C and radius a has a spherically symmetric electric charge
density. The total charge of the object is Q > 0.
a. Determine the magnitude and direction of the electric field at point P, which is a distance R > a to the
right of the sphere’s center.
b. Determine the flux of the electric field through the spherical surface centered at C and passing through
P.
A point particle of charge -Q is now placed a distance R below point P. as shown above.
c. Determine the magnitude and direction of the electric field at point P.
d. Now consider four point charges, \(q_1,q_2, q_3\), and \(q_4\), that lie in the the plane of the page as shown in the
diagram above. Imagine a three-dimensional closed surface whose cross section in the plane of the
page is indicated.
i. Which of these charges contribute to the net electric flux through the surface?
ii. Which of these charges contribute to the electric field at point \(P_1\)?
iii. Are your answers to i and ii the same or are they different? Explain why this is so.
e. If the net charge enclosed by a surface is zero, does this mean that the field is zero at all points on the
surface? Justify your answer.
f. If the field is zero at all points on a surface, does this mean there is no net charge enclosed by the
surface? Justify your answer.
Answer/Explanation
Ans:
a. The sphere can be treated as a point charge: \(E=kQ/R^2\) to the right
b. The flux is represented by either side of Gauss’s Law: \(Q/\varepsilon_0\) or \(4 \pi R^2 E\)
c. The new field is the vector sum of the fields from the sphere and the point charge
From the sphere: \(kQ/R^2\) to the right form the point charge: \(kQ/R^2\) downward from the Pythagoras theorem: \(E_{net}=\sqrt{2}kQ/R^2\) \(45^0\) form the horizontal, down and to the right
d. Only those charges inside contribution to the net electric flux: \(q_2\) and \(q_3\)
ii. All four charges contribute to the value of the electric field at point \(P_1\)
iii. iii. Different, the electric field is the sum of the individual fields while the flux is the sum of components over
the whole surface
e. No. Charges may exist outside the surface that would contribute to a field on the surface or, for example, the
surface may enclose a dipole, which would cause a net field at some points
f. Yes. A zero net field means the flux is also zero. Since charges inside the surface contribute to the net flux,
there can be no net charge inside the surface.