Question 1 of 7: Two Point Charges — Field, Cancelling Charge, Force and Work

Question 1: Two Point Charges — Field, Cancelling Charge, Force and Work (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) $\vec E$ at $P_1$. (b) the charge $Q$ at $P_2$ whose own field cancels that $E$ at $P_1$. (c) the force on $Q$. (d) work done moving $Q$ from $P_2$ to $P_1$.

Approach. Superpose Coulomb fields from $q_1,q_2$ at $P_1$ (part a); require the field of an unknown $Q$ at $P_2$ to cancel it (part b); use $\vec F=Q\vec E$ for the force on $Q$ once it sits at $P_2$ (part c); and use $W_{\text{ext}}=Q(V_{P_1}-V_{P_2})$, with $V$ due to $q_1,q_2$ only, for the work of part (d).

1. Part (a) — field at $P_1$. Each charge is a distance $r=\sqrt{4^2+4^2}=5.657$ cm $=0.05657$ m from $P_1$, at $45^\circ$ to the axes, so the $y$-components of $\vec E_1$ (from $q_1$) and $\vec E_2$ (from $q_2$) are equal and opposite: $$E_{\text{each}}=\frac{kq}{r^2}=\frac{(8.988\times10^9)(2\times10^{-9})}{(0.05657)^2}=5617\ \text{V/m}$$ Both contributions have the same $x$-component ($-\cos45^\circ$ of $E_{\text{each}}$), which add: $$\boxed{\vec E_{P_1}=(-7944,\,0,\,0)\ \text{V/m},\ |\vec E_{P_1}|=7944\ \text{V/m, in the}-x\text{ direction}}$$
2. Part (b) — cancelling charge $Q$ at $P_2$. $P_2$ is $r_{QP_1}=8$ cm $=0.08$ m from $P_1$, directly along the $x$-axis, so the field $Q$ produces at $P_1$ is purely $x$-directed. To cancel $\vec E_{P_1}=(-7944,0,0)$ V/m, $Q$'s own field at $P_1$ must point in $+x$ — since $P_1$ lies in the $-x$ direction from $Q$'s position $P_2$, a field pointing back toward $+x$ (i.e. toward $Q$) requires $Q$ to be negative: $$\frac{k|Q|}{r_{QP_1}^2}=7944 \implies |Q|=\frac{7944\times(0.08)^2}{8.988\times10^9}$$ $$\boxed{Q=-5.657\ \text{nC}}$$
3. Part (c) — force on $Q$. With $Q$ now physically at $P_2$, find the field $q_1,q_2$ produce there (by the same symmetry as part a, but mirrored): $\vec E_{P_2}=(+7944,0,0)$ V/m. The force on the negative charge is opposite to this field: $$\vec F=Q\vec E_{P_2}=(-5.657\times10^{-9})(7944,0,0)$$ $$\boxed{\vec F=-44.94\ \mu\text{N in }x,\ |\vec F|=44.94\ \mu\text{N, in the}-x\text{ direction}}$$ (i.e. $Q$ is pulled toward the two positive charges, as expected for an attractive force.)
4. Part (d) — work moving $Q$ from $P_2$ to $P_1$. The work done by an external force against the field of $q_1,q_2$ is $W_{\text{ext}}=Q(V_{P_1}-V_{P_2})$, using the potential due to $q_1,q_2$ only (a charge does no work against its own field). Because $q_1,q_2$ sit symmetrically on the $y$-axis, and $P_1,P_2$ are mirror-image points across that same axis, every distance from $q_1$ or $q_2$ to $P_1$ equals the corresponding distance to $P_2$ — so $V_{P_1}=V_{P_2}$ exactly: $$V_{P_1}=V_{P_2}=\frac{2kq_1}{r}=\frac{2(8.988\times10^9)(2\times10^{-9})}{0.05657}=635.6\ \text{V}$$ $$\boxed{W_{\text{ext}}=Q(V_{P_1}-V_{P_2})=0\ \text{J}}$$ The mover does zero net work — $P_1$ and $P_2$ lie on the same equipotential of the $q_1,q_2$ pair.