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💡AP Physics C: Electricity and Magnetism
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💡AP Physics C: Electricity and Magnetism

FRQ 4 – Qualitative/Quantitative Translation
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Unit 8: Electric Charges, Fields, and Gauss's Law
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Practice FRQ 1 of 201/20

4. A point charge +q=3.0 μC+q = 3.0\,\mu\text{C}+q=3.0μC is held fixed on the zzz-axis at z=+0.20 mz = +0.20\,\text{m}z=+0.20m above a large, flat boundary surface that lies in the xyxyxy-plane at z=0z=0z=0, as shown in Figure 1. The region z>0z>0z>0 is air with permittivity ε0\varepsilon_0ε0​. The region z<0z<0z<0 is filled with a linear dielectric material with permittivity ε=4ε0\varepsilon = 4\varepsilon_0ε=4ε0​. Ignore edge effects and assume the boundary is infinite. A small test charge +q0=2.0 nC+q_0 = 2.0\,\text{nC}+q0​=2.0nC with mass m=1.0 gm = 1.0\,\text{g}m=1.0g is placed at point PPP on the zzz-axis at z=+0.60 mz = +0.60\,\text{m}z=+0.60m. The gravitational field is g=9.8 m/s2g = 9.8\,\text{m/s}^2g=9.8m/s2 downward.

Figure 1. Point charge above an infinite planar boundary between air (ε0) and a dielectric (ε = 4ε0), with a test charge at point P on the z-axis.

Figure 1
A.

Let FEF_EFE​ be the magnitude of the electric force on the test charge +q0+q_0+q0​ at point PPP due to the source charge and induced charges at the boundary. Let FgF_gFg​ be the magnitude of the gravitational force on the test charge.

Indicate whether FEF_EFE​ is greater than, less than, or equal to FgF_gFg​ by writing one of the following.

  • FE>FgF_E > F_gFE​>Fg​
  • FE<FgF_E < F_gFE​<Fg​
  • FE=FgF_E = F_gFE​=Fg​

Justify your answer by identifying the direction of the electric force at PPP and comparing the expected magnitudes using relevant physics relationships.

B.

Derive an expression for the magnitude EPE_PEP​ of the electric field at point PPP in the air region. Use the method of images for a point charge above a planar interface between two linear media with permittivities ε1\varepsilon_1ε1​ (for z>0z>0z>0) and ε2\varepsilon_2ε2​ (for z<0z<0z<0).

Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Your final expression should be in terms of qqq, ε1\varepsilon_1ε1​, ε2\varepsilon_2ε2​, the Coulomb constant k=14πε0k = \frac{1}{4\pi\varepsilon_0}k=4πε0​1​, and the distances from PPP to the real charge and to the image charge. Clearly indicate any superposition you use.

(For this configuration, the image charge magnitude is q′=(ε1−ε2ε1+ε2)qq' = \left(\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1+\varepsilon_2}\right)qq′=(ε1​+ε2​ε1​−ε2​​)q located the same distance below the boundary as the real charge is above it.)

Figure 2. Same charge locations as Figure 1, but the lower half-space is a conductor with a conducting plane at z = 0.

Figure 2
C.

Indicate whether EcondE_{\text{cond}}Econd​ is greater than, less than, or equal to EPE_PEP​ by writing one of the following. Later, the dielectric region is replaced by a conductor, as shown in Figure 2. The charge +q=3.0 μC+q = 3.0\,\mu\text{C}+q=3.0μC remains fixed at z=+0.20 mz = +0.20\,\text{m}z=+0.20m, and point PPP remains at z=+0.60 mz = +0.60\,\text{m}z=+0.60m in air. Let EcondE_{\text{cond}}Econd​ be the magnitude of the electric field at PPP for the conductor case, and let EPE_PEP​ be the magnitude from part B for the dielectric case with ε2=4ε1\varepsilon_2 = 4\varepsilon_1ε2​=4ε1​.

  • Econd>EPE_{\text{cond}} > E_PEcond​>EP​
  • Econd<EPE_{\text{cond}} < E_PEcond​<EP​
  • Econd=EPE_{\text{cond}} = E_PEcond​=EP​

Briefly justify your answer by referencing your expression from part B and the appropriate image-charge factor for a conducting plane.

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Free Response Question Practice

This practice environment simulates the AP AP Physics C: Electricity and Magnetism Free Response Questions section. Here are some guidelines:

  • Read each question carefullybefore responding. Pay attention to command verbs like "identify," "explain," "analyze," or "evaluate."
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Tip: Answer all parts of each question. Partial credit is often available, so even if you are unsure, provide what you know.