AP Physics 2

Unit 10

Electric Force, Field, and Potential

8 practice questions · Free response

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Notes

Charge

  • Charge comes in positive and negative; like charges repel, opposites attract. Charge is conserved and quantized in units of $e = 1.6 \times 10^{-19}$ C.
  • Objects charge by friction, conduction (contact), or induction (a nearby charge shifts electrons, then grounding).
  • Conductors let charge move freely; insulators don't. Neutral objects can be attracted to charged ones because they polarize.

Coulomb's law

$$F = k\frac{|q_1q_2|}{r^2}, \quad k = 9.0 \times 10^9\ \text{N·m}^2/\text{C}^2$$

Doubling the distance makes the force $\displaystyle \frac{1}{4}$ as big.

Electric field

  • $\displaystyle \vec E = \frac{\vec F}{q}$ (N/C). For a point charge, $\displaystyle E = k\frac{|q|}{r^2}$.
  • Field lines point away from positive and toward negative charges. Closer lines mean a stronger field.
  • A positive charge feels a force along the field; a negative charge feels a force opposite the field.
  • Between parallel plates, the field is uniform: $\displaystyle E = \frac{\Delta V}{d}$.

Electric potential and potential energy

  • Potential energy of two charges: $\displaystyle U_E = k\frac{q_1q_2}{r}$.
  • Electric potential $\displaystyle V = \frac{U}{q}$ (volts). For a point charge, $\displaystyle V = k\frac{q}{r}$; potentials add as scalars.
  • Work to move a charge: $W = q\Delta V$. Positive charges naturally move toward lower potential.
  • Equipotential lines are perpendicular to field lines. Moving along one takes no work.

Practice

Two charges are 2.0 m apart. If the distance becomes 4.0 m, the force between them is:
What is the force between two 1.0 μC charges 0.30 m apart?
Electric field lines near a negative point charge:
An electron is placed in a field pointing east. The force on it points:
A charged balloon sticks to a neutral wall because:
How much work is needed to move a 2.0 C charge through a potential difference of 12 V?
Parallel plates 0.010 m apart have a potential difference of 100 V. What is the field between them?
Moving a charge along an equipotential line requires:

Free response

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Free-response question

Two charges, $+4.0$ μC at $x = 0$ and $+1.0$ μC at $x = 3.0$ m, are on the $x$-axis.

(a) Find the magnitude of the force between them.

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$\displaystyle F = \frac{(9.0 \times 10^9)(4.0 \times 10^{-6})(1.0 \times 10^{-6})}{9.0} = 4.0 \times 10^{-3}$ N.

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(b) Find the point between them where the electric field is zero.

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$\displaystyle \frac{4.0}{x^2} = \frac{1.0}{(3.0 - x)^2} \Rightarrow \frac{2}{x} = \frac{1}{3 - x} \Rightarrow x = 2.0$ m.

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(c) Is the electric potential zero at that point? Explain.

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No. Both charges are positive, so both contribute positive potential: $\displaystyle V = k\left(\frac{4.0 \times 10^{-6}}{2.0} + \frac{1.0 \times 10^{-6}}{1.0}\right) = 2.7 \times 10^4$ V.

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