Electric Forces and Fields — Hard Practice Quiz
A Physics cheat sheet for Electric Forces and Fields — every key formula with its symbols defined — plus a hard-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 \(r\).
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 \(q_0\).
Electric Field of a Point Charge: Magnitude of electric field \(E\) at distance \(r\) from a point charge \(q\).
Electric Flux: Flux \(\Phi_E\) through a surface of area \(A\) where electric field \(E\) makes angle \(\phi\) with the normal.
Gauss' Law: Net electric flux \(\Phi_E\) through a closed surface equals net enclosed charge \(Q\) divided by permittivity of free space \(\epsilon_0\).
Where: \(\epsilon_0\) = permittivity of free space \((8.85 \times 10^{-12} C^2/(N\cdot m^2))\)
Practice quiz
A point charge $q_1$ creates an electric field. A second point charge $q_2$ is placed at a distance $r$ from $q_1$, experiencing a force of magnitude $F$. If $q_2$ is then replaced by a test charge $q_0$ at the same distance $r$, what is the magnitude of the electric field $E$ at that point in terms of $F$, $q_1$, and $q_2$?
- $\frac{F}{q_0}$
- $\frac{F}{q_2}$
- $\frac{F q_0}{q_2}$
- $\frac{F q_2}{q_0}$
Answer: $\frac{F}{q_2}$
A point charge $Q$ is located at the center of a cube with side length $L$. What is the magnitude of the electric flux through one face of the cube?
- $\frac{Q}{\epsilon_0}$
- $\frac{Q}{6\epsilon_0}$
- $\frac{Q L^2}{\epsilon_0}$
- $\frac{Q}{L^2 \epsilon_0}$
Answer: $\frac{Q}{6\epsilon_0}$
Two point charges, $q_A = +2q$ and $q_B = -q$, are separated by a distance $d$. At what distance from $q_B$ (along the line connecting them or its extension) is the net electric field zero?
- $d(\sqrt{2}-1)$
- $d(\sqrt{2}+1)$
- $\frac{d}{\sqrt{2}}$
- $\frac{d}{2}$
Answer: $d(\sqrt{2}+1)$
A uniform electric field of magnitude $E$ passes through a circular disk of radius $R$. If the electric field lines are parallel to the plane of the disk, what is the electric flux through the disk?
- $E \pi R^2$
- $\frac{E \pi R^2}{2}$
- $0$
- $E \pi R^2 \cos(45^\circ)$
Answer: $0$
A spherical Gaussian surface encloses a net charge $Q$. If the radius of the Gaussian surface is doubled, and simultaneously, the enclosed charge is halved, how does the net electric flux through the surface change?
- It doubles.
- It is halved.
- It remains the same.
- It is quartered.
Answer: It is halved.
An electric field $E$ is produced by a point charge $q$. If a test charge $q_0$ is placed at a distance $r$ from $q$, it experiences a force $F$. If the distance $r$ is doubled, and the magnitude of $q$ is also doubled, how does the force $F$ on $q_0$ change?
- It doubles.
- It is halved.
- It remains the same.
- It is quartered.
Answer: It is halved.
A point charge $q$ is placed at the origin. A spherical Gaussian surface of radius $R$ is centered at $x=2R$ on the x-axis. What is the net electric flux through this Gaussian surface?
- $\frac{q}{\epsilon_0}$
- $\frac{q}{2\epsilon_0}$
- $0$
- $\frac{q}{4\pi R^2 \epsilon_0}$
Answer: $0$
A uniform electric field of magnitude $E = 100 \text{ N/C}$ passes through a flat square surface of side length $2.0 \text{ m}$. If the electric flux through the surface is $100 \text{ N} \cdot \text{m}^2/\text{C}$, what is the angle $\phi$ between the electric field vector and the normal to the surface?
- $\arccos(1)$
- $\arccos(0.5)$
- $\arccos(0.25)$
- $\arccos(0)$
Answer: $\arccos(0.25)$
Two identical positive point charges, each of magnitude $q$, are separated by a distance $2r$. What is the magnitude of the net electric field at the midpoint between them?
- $0$
- $k \frac{q}{r^2}$
- $2k \frac{q}{r^2}$
- $k \frac{q}{(2r)^2}$
Answer: $0$
A charge $Q$ is uniformly distributed over a very large, thin, non-conducting sheet, creating a uniform electric field perpendicular to the sheet with magnitude $E = \frac{\sigma}{2\epsilon_0}$, where $\sigma$ is the surface charge density. If a point charge $q_0$ is placed at a distance $d$ from this sheet, what is the magnitude of the force it experiences?
- $q_0 \frac{\sigma}{2\epsilon_0}$
- $q_0 \frac{\sigma}{2\epsilon_0 d^2}$
- $\frac{\sigma}{2\epsilon_0}$
- $\frac{Q q_0}{4\pi \epsilon_0 d^2}$
Answer: $q_0 \frac{\sigma}{2\epsilon_0}$
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