Electric Forces and Fields — Practice Quiz

A Physics cheat sheet for Electric Forces and Fields — every key formula with its symbols defined — plus a medium-level practice quiz to test recall.

Formulas & key concepts

Coulomb's Law: Magnitude of electrostatic force \(F\) between two point charges \(q_1\) and \(q_2\) separated by distance \(r\).

$$F = k \frac{|q_1 q_2|}{r^2}$$

Where: \(k\) = Coulomb constant \((8.99 \times 10^9 N\cdot m^2/C^2)\)

Electric Field Definition: Electric field \(\vec{E}\) is the electric force \(\vec{F}\) experienced by a small test charge \(q_0\) divided by \(q_0\).

$$\vec{E} = \frac{\vec{F}}{q_0}$$

Electric Field of a Point Charge: Magnitude of electric field \(E\) at distance \(r\) from a point charge \(q\).

$$E = k \frac{|q|}{r^2}$$

Electric Flux: Flux \(\Phi_E\) through a surface of area \(A\) where electric field \(E\) makes angle \(\phi\) with the normal.

$$\Phi_E = (E \cos \phi) A$$

Gauss' Law: Net electric flux \(\Phi_E\) through a closed surface equals net enclosed charge \(Q\) divided by permittivity of free space \(\epsilon_0\).

$$\Phi_E = \frac{Q}{\epsilon_0}$$

Where: \(\epsilon_0\) = permittivity of free space \((8.85 \times 10^{-12} C^2/(N\cdot m^2))\)

Practice quiz

  1. If the distance $r$ between two point charges is doubled, how does the magnitude of the electrostatic force $F$ between them change?

    • A. It is quadrupled.
    • B. It is halved.
    • C. It is quartered.
    • D. It remains the same.

    Answer: C. It is quartered.

  2. Two point charges, $q_1 = +2.0 \times 10^{-6} \text{ C}$ and $q_2 = -3.0 \times 10^{-6} \text{ C}$, are separated by a distance of $0.50 \text{ m}$. What is the magnitude of the electrostatic force between them? Use $k = 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$.

    • A. $0.108 \text{ N}$
    • B. $0.216 \text{ N}$
    • C. $0.432 \text{ N}$
    • D. $0.054 \text{ N}$

    Answer: B. $0.216 \text{ N}$

  3. A positive test charge $q_0$ experiences an electric force $\vec{F}$ directed to the right. In what direction is the electric field $\vec{E}$ at the location of the test charge?

    • A. To the right.
    • B. To the left.
    • C. Upwards.
    • D. Downwards.

    Answer: A. To the right.

  4. What is the magnitude of the electric field at a distance of $0.10 \text{ m}$ from a point charge of $q = +5.0 \times 10^{-9} \text{ C}$? Use $k = 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$.

    • A. $4.5 \times 10^2 \text{ N/C}$
    • B. $4.5 \times 10^3 \text{ N/C}$
    • C. $9.0 \times 10^3 \text{ N/C}$
    • D. $9.0 \times 10^2 \text{ N/C}$

    Answer: B. $4.5 \times 10^3 \text{ N/C}$

  5. If the distance $r$ from a point charge is tripled, how does the magnitude of the electric field $E$ at that point change?

    • A. It becomes one-third.
    • B. It becomes one-ninth.
    • C. It becomes three times.
    • D. It becomes nine times.

    Answer: B. It becomes one-ninth.

  6. For a uniform electric field $E$ passing through a flat surface of area $A$, when is the electric flux $\Phi_E$ through the surface zero?

    • A. When the electric field is parallel to the surface.
    • B. When the electric field is perpendicular to the surface.
    • C. When the electric field is at a $45^\circ$ angle to the surface.
    • D. When the electric field is zero.

    Answer: A. When the electric field is parallel to the surface.

  7. A uniform electric field of magnitude $E = 200 \text{ N/C}$ passes through a flat surface of area $A = 0.50 \text{ m}^2$. If the electric field makes an angle of $60^\circ$ with the normal to the surface, what is the electric flux through the surface?

    • A. $100 \text{ N} \cdot \text{m}^2/\text{C}$
    • B. $50 \text{ N} \cdot \text{m}^2/\text{C}$
    • C. $86.6 \text{ N} \cdot \text{m}^2/\text{C}$
    • D. $0 \text{ N} \cdot \text{m}^2/\text{C}$

    Answer: B. $50 \text{ N} \cdot \text{m}^2/\text{C}$

  8. According to Gauss' Law, the net electric flux $\Phi_E$ through any closed surface is directly proportional to what quantity?

    • A. The total charge outside the surface.
    • B. The total charge enclosed within the surface.
    • C. The area of the closed surface.
    • D. The electric field strength at the surface.

    Answer: B. The total charge enclosed within the surface.

  9. A closed surface encloses a net electric flux of $\Phi_E = 5.0 \times 10^3 \text{ N} \cdot \text{m}^2/\text{C}$. What is the net charge $Q$ enclosed within the surface? Use $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)$.

    • A. $4.43 \times 10^{-8} \text{ C}$
    • B. $5.65 \times 10^{14} \text{ C}$
    • C. $1.77 \times 10^{-7} \text{ C}$
    • D. $8.85 \times 10^{-12} \text{ C}$

    Answer: A. $4.43 \times 10^{-8} \text{ C}$

  10. An electric field of magnitude $E = 1.5 \times 10^4 \text{ N/C}$ exists at a certain point. If a charge of $q = -2.0 \times 10^{-9} \text{ C}$ is placed at this point, what is the magnitude of the electric force it experiences?

    • A. $3.0 \times 10^{-5} \text{ N}$
    • B. $7.5 \times 10^{-14} \text{ N}$
    • C. $1.3 \times 10^{13} \text{ N}$
    • D. $3.0 \times 10^{-9} \text{ N}$

    Answer: A. $3.0 \times 10^{-5} \text{ N}$

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