AP Physics B : Understanding Electrostatics

Study concepts, example questions & explanations for AP Physics B

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Example Questions

Example Question #1 : Ap Physics 2

Charges A and B are placed a distance of \displaystyle 1.5m from one another. The charge of particle A is \displaystyle 0.003C whereas the charge of particle B is \displaystyle 0.006C. Charge B experiences an electrostatic force of \displaystyle F_b from charge A. Similarly, charge A experiences an electrostatic force of \displaystyle F_a from charge B.

A student places these two charges in a vacuum medium and measures \displaystyle F_a. A second student places these two charges in a glass medium and measures \displaystyle F_a. What is ratio of the force measured in the glass medium to the force measured in the vacuum?

The dielectric constant of glass is \displaystyle 5.

Possible Answers:

\displaystyle 5

\displaystyle 2.5

\displaystyle 0.2

\displaystyle 0.5

Correct answer:

\displaystyle 0.2

Explanation:

You can simplify this question tremendously by using the definition of a dielectric constant. Dielectric constant is defined as the ratio of the electrostatic force in vacuum to the electrostatic force in the medium (in this case glass).

\displaystyle k=\frac{F_{e,vac}}{F_{e,glass}}=5

The question is asking for the reciprocal of this value: the ratio of the force in glass to the force in the vacuum. Our answer is calculated by taking the reciprocal of the dielectric constant of glass.

\displaystyle \frac{1}{k}=\frac{F_{e,glass}}{F_{e,vac}}=\frac{1}{5}=0.2

 

Example Question #1 : Ap Physics 2

How much work is done by the electric field moving an electron along an equipotential surface with a potential of \displaystyle 3V?

Possible Answers:

\displaystyle 0J

\displaystyle 2.42J

\displaystyle 5.33J

We must know the strength of the electric field to solve

Correct answer:

\displaystyle 0J

Explanation:

Moving a charge along an equipotential surface will involve no work done by the electric field. The potential is constant throughout an equipotential surface; therefore, the potential difference experienced by the electron will be zero. Remember that energy is dependent on the potential difference.

\displaystyle W=q\Delta V=q(E*d)

If the potential difference is zero then the energy (and work) will be zero.

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