The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 21 — Electric Charge & the Electric Field

Problems with worked solutions · 12 points · 3 extra credit

Chapter 21 — Electric Charge & the Electric Field

Core: 12 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

Use $k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$.

1. 1. (2 pts ••) Two small spheres carry $q_1=+4.0\,\mu\text{C}$ and $q_2=-7.0\,\mu\text{C}$ and are $30\,\text{cm}$ apart.

   (a) Find the magnitude of the force each exerts on the other.

   (b) A third charge $q_3=+2.0\,\mu\text{C}$ is placed at the midpoint between them. Find the magnitude and direction of the net force on $q_3$.

   Solution

   1.

   (a) $F=\dfrac{k|q_1 q_2|}{r^2}=\dfrac{(8.99\times10^{9})(4.0\times10^{-6})(7.0\times10^{-6})}{(0.30)^2}\approx 2.8\,\text{N}$ (attractive).

   (b) Each charge is $0.15\,\text{m}$ from $q_3$. $q_1$ repels $q_3$ toward $q_2$; $q_2$ attracts $q_3$ toward $q_2$ — same direction. $F_{13}=\dfrac{k(4.0\times10^{-6})(2.0\times10^{-6})}{(0.15)^2}=3.20\,\text{N}$ and $F_{23}=\dfrac{k(7.0\times10^{-6})(2.0\times10^{-6})}{(0.15)^2}=5.59\,\text{N}$. Net $=8.8\,\text{N}$ toward $q_2$.

2. 2. (3 pts •••) Four equal charges $+q$ sit at the corners of a square of side $L$.

   (a) Find the magnitude and direction of the net force on one corner charge.

   (b) Find the electric field at the center of the square.

   (c) A charge $-Q$ is placed at the center. Is it in equilibrium? Is that equilibrium stable? Justify your answer.

   Solution

   2.

   (a) The two adjacent charges (distance $L$) each contribute $kq^2/L^2$; along the diagonal their sum is $\sqrt{2}\,kq^2/L^2$. The far corner (distance $\sqrt{2}\,L$) adds $kq^2/2L^2$ along the same diagonal. Total $=\dfrac{kq^2}{L^2}\left(\sqrt{2}+\tfrac12\right)\approx 1.91\,\dfrac{kq^2}{L^2}$, outward along the diagonal.

   (b) By symmetry the four equal contributions cancel: $E=0$ at the center.

   (c) The net force on $-Q$ is zero (the field is zero there), so it is in equilibrium — but the equilibrium is unstable. Nudge $-Q$ slightly toward one corner: that charge is now closer and pulls harder than the far corner, so the net force drives $-Q$ farther from the center rather than restoring it.

3. 3. (2 pts ••) A charge $q_1=+5.0\,\text{nC}$ is at the origin and $q_2=-3.0\,\text{nC}$ is at $(4.0\,\text{cm},\,0)$.

   (a) Find the electric field at the point $P=(0,\,3.0\,\text{cm})$ (magnitude and direction).

   (b) Find the force on a $-1.0\,\text{nC}$ charge placed at $P$.

   Solution

   3.

   (a) From $q_1$: $E_1=\dfrac{k(5.0\times10^{-9})}{(0.030)^2}\approx 5.0\times10^{4}\,\text{N/C}$, pointing $+y$. From $q_2$: $r_2=\sqrt{0.04^2+0.03^2}=0.050\,\text{m}$, $E_2=\dfrac{k(3.0\times10^{-9})}{(0.050)^2}=1.08\times10^{4}\,\text{N/C}$ toward $q_2$, direction $(0.8,-0.6)$. Sum $\mathbf{E}\approx(8.6\times10^{3},\,4.35\times10^{4})\,\text{N/C}$; magnitude $\approx 4.4\times10^{4}\,\text{N/C}$, about $79^\circ$ above $+x$.

   (b) $F=|q|E=(1.0\times10^{-9})(4.4\times10^{4})\approx 4.4\times10^{-5}\,\text{N}$, opposite $\mathbf{E}$ (the charge is negative).

4. 4. (2 pts ••) An electric dipole has charges $\pm q$ with $q=3.0\,\text{nC}$ separated by $d=2.0\,\text{mm}$, in a uniform field $E=4.0\times10^{5}\,\text{N/C}$.

   (a) Find the maximum torque on the dipole.

   (b) Find the work needed to rotate it from aligned ($\theta=0$) to $\theta=90^\circ$.

   (c) …and from $\theta=0$ to $\theta=180^\circ$.

   Solution

   4. $p=qd=(3.0\times10^{-9})(2.0\times10^{-3})=6.0\times10^{-12}\,\text{C}\cdot\text{m}$.

   (a) $\tau_{\max}=pE=2.4\times10^{-6}\,\text{N}\cdot\text{m}$.

   (b) $U(\theta)=-pE\cos\theta$, so $W=U(90^\circ)-U(0)=pE=2.4\times10^{-6}\,\text{J}$.

   (c) $W=U(180^\circ)-U(0)=2pE=4.8\times10^{-6}\,\text{J}$.

5. 5. (2 pts ••) In a simplified inkjet printer, a droplet of mass $m=1.3\times10^{-10}\,\text{kg}$ and charge $q=-1.5\times10^{-13}\,\text{C}$ enters a $1.6\,\text{cm}$-long region of uniform field $E=8.0\times10^{4}\,\text{N/C}$ (pointing down) at a horizontal speed of $20\,\text{m/s}$. Ignoring gravity, find the vertical displacement of the droplet while it travels through the field region.

   Solution

   5. $|F|=|q|E=(1.5\times10^{-13})(8.0\times10^{4})=1.2\times10^{-8}\,\text{N}$ (upward, since $q<0$). Then $a=F/m=92\,\text{m/s}^2$ and $t=L/v=0.016/20=8.0\times10^{-4}\,\text{s}$, giving $y=\tfrac12 a t^2\approx 3.0\times10^{-5}\,\text{m}$ (upward).

6. 6. (1 pt •) A charged particle feels an upward force of $4.0\times10^{-6}\,\text{N}$ in a uniform field of $2.0\times10^{3}\,\text{N/C}$ that points downward. Find the sign and magnitude of its charge.

   Solution

   6. $|q|=F/E=\dfrac{4.0\times10^{-6}}{2.0\times10^{3}}=2.0\times10^{-9}\,\text{C}$. The force is up while the field points down, so the charge is negative: $q=-2.0\,\text{nC}$.

7. Extra Credit (3 pts •••)

   A thin nonconducting rod of length $L$ carries uniform linear charge density $\lambda$.

   (a) Starting from Coulomb's law, integrate to find the electric field at a point on the rod's axis a distance $a$ beyond the near end.

   (b) Do the same for a point a perpendicular distance $y$ from the rod's midpoint, and show your result reduces to the infinite-line field $E=\dfrac{\lambda}{2\pi\varepsilon_0 y}$ as $L\to\infty$.

   (c) Show that in the appropriate far-field limit each result reduces to that of a point charge.

   Solution

   Extra Credit

   (a) With the near end a distance $a$ away, an element $dq=\lambda\,dx$ at distance $x$ contributes $dE=k\lambda\,dx/x^2$ along the axis:

   $$E=\int_{a}^{a+L}\frac{k\lambda}{x^2}\,dx=k\lambda\left(\frac{1}{a}-\frac{1}{a+L}\right)=\frac{k\lambda L}{a(a+L)}.$$

   (b) On the perpendicular bisector only the perpendicular component survives:

   $$E_y=k\lambda y\int_{-L/2}^{L/2}\frac{dx}{(x^2+y^2)^{3/2}}=\frac{k\lambda L}{y\sqrt{y^2+(L/2)^2}}.$$

   As $L\to\infty$, $\sqrt{y^2+(L/2)^2}\to L/2$, so $E_y\to\dfrac{2k\lambda}{y}=\dfrac{\lambda}{2\pi\varepsilon_0 y}$.

   (c) Far field with $Q=\lambda L$: part (a) for $a\gg L$ gives $E\to kQ/a^2$; part (b) for $y\gg L$ gives $E_y\to kQ/y^2$ — the point-charge result.