Essential Physics

Section 4 review

Electric potential energy refers to the energy stored in a charged particle when you move that particle into a particular location. Electric potential is the electric potential energy per unit charge, and it has units of volts (one volt being one joule per coulomb). Equipotential lines trace regions of equal electric potential, much as contour lines trace regions of equal elevation on topographic maps. A capacitor is an electrical component that can store a certain amount of electric charge per volt applied to it; the amount of stored charge is proportional to the applied voltage and the capacitance. Read the text aloud

Read the text aloud

electric potential, equipotential, electric potential energy, capacitor, capacitance

V = \frac{E_{p}}{q}

V = \frac{k_{e} q}{r}

E_{p} = \frac{k_{e} q_{1} q_{2}}{r}

q = C V

E_{p} = \frac{1}{2} C V^{2}

C = \frac{ε A}{d}

C_{e q} = C_{1} + C_{2}

\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

Review problems and questions

Which diagram might match a fully charged 1.5 V battery?

Which of the following diagrams could be applied to a fully charged 1.5 V battery? (More than one selection may be correct.)

Electric force field around a charged particle

Charge 1 (with q₁ coulombs) generates the electric field shown here. Charge 2 (with q₂ coulombs) is brought to a distance r from Charge 1.
1. How would doubling q₁ affect the electric potential energy E_p?
2. What about doubling q₂?
3. How about doubling both charge values?
4. Finally, what about doubling the distance, r?

Answer:

Doubling q₁ will double the electric potential energy E_p.
Doubling q₂ will likewise double the electric potential energy E_p.
Doubling both charge values will quadruple the electric potential energy.
Doubling the separation r of the two charges will halve the electric potential energy.

Solution:

Doubling q₁ will double the electric potential energy E_p. This can be seen by referring to the formula $\begin{matrix} old E_{p} = \frac{k_{e} q_{1} q_{2}}{r} \\ new E_{p} = \frac{k_{e} (2 q_{1}) q_{2}}{r} = \frac{2 k_{e} (q_{1}) q_{2}}{r} = 2 (old E_{p}) \end{matrix}$
Doubling q₂ will likewise double the electric potential energy E_p.
Doubling both charge values will quadruple the electric potential energy, because the energy is directly proportional to both charge q₁ and q₂.
Doubling the separation r of the two charges will halve the electric potential energy: $\begin{matrix} old E_{p} = \frac{k_{e} q_{1} q_{2}}{r} \\ new E_{p} = \frac{k_{e} q_{1} q_{2}}{2 r} = \frac{1}{2} (old E_{p}) \end{matrix}$

Two circuits with capacitors, one in parallel and the other in series

Two circuits are built with identical components, but one has the pair of microfarad capacitors in parallel, while the other has them in series. Which will be able to build up a greater charge?

The parallel circuit will be able to build up a greater charge after the switches of both circuits are closed. That is because the charge that can accumulate in a capacitor network is proportional to its equivalent capacitance, measured in farads (or microfarads, as is the case here). The equivalent capacitance of a parallel circuit is given by the formula

C_{e q} = C_{1} + C_{2}

In this case, the equivalent capacitance is simply 2 μF. In the series case, the equivalent capacitance is given by the more complex formula

\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

Substituting 1 μf for each of C₁ and C₂ yields

\frac{1}{C_{e q}} = \frac{1}{1 μF} + \frac{1}{1 μF} = \frac{2}{1 μF}

Inverting both sides of the equation then shows that C_eq = ½ μF. This shows that the parallel circuit therefore has four times as much capacitance as the series circuit and will accumulate four times the charge (since the voltage sources are identical).

Electic field lines between parallel charged plates

The positively (top) and negatively (bottom) charged plates of a parallel plate capacitor are shown here, along with the electric field between them. Draw several equipotential lines for the region between the plates. Indicate which has the highest electric potential.


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