Electric Field and Force from Two Point Charges: 04-BS-9, December 2019, Q1

Question 1: Electric Field and Force from Two Point Charges (20 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.

Quantity	Value
Charge $q_1$, position	$6.40$ nC at $(0,4,0)$ cm
Charge $q_2$, position	$14.23$ nC at $(2,-3,0)$ cm
Field/observation point $P$	$(-4,0,0)$ cm
Charge $Q$ at $P$ (part d)	$-20$ nC

Find. (a) $\vec E_1$ at $P$ due to $q_1$. (b) $\vec E_2$ at $P$ due to $q_2$. (c) the total $\vec E$ at $P$. (d) the total electrostatic force on $Q$ placed at $P$.

Charges q1, q2 and observation point P in the z=0 plane (cm coordinates). E1 and E2 point away from their respective positive source charges.

Approach. Apply Coulomb's law $\vec E=kq\hat r/r^2$ for each source charge in turn, superpose the two vectors for the total field, then use $\vec F=Q\vec E$ for the force on the test charge.

Part (a) — field of $q_1$ at $P$. Displacement $\vec r_1=P-q_1=(-4-0,\,0-4,\,0)\text{ cm}=(-0.04,-0.04,0)$ m, $r_1=|\vec r_1|=0.05657$ m: $$\vec E_1=\frac{kq_1}{r_1^2}\hat r_1=\frac{(8.988\times10^9)(6.40\times10^{-9})}{(0.05657)^2}\hat r_1$$ $$\boxed{\vec E_1=(-1.271\times10^4,\,-1.271\times10^4,\,0)\ \text{V/m},\ |\vec E_1|=1.798\times10^4\ \text{V/m}}$$ directed at $225^\circ$ from $+x$ (into the third quadrant, $45^\circ$ below $-x$).
Part (b) — field of $q_2$ at $P$. $\vec r_2=P-q_2=(-4-2,\,0-(-3),\,0)\text{ cm}=(-0.06,0.03,0)$ m, $r_2=0.06708$ m: $$\vec E_2=\frac{kq_2}{r_2^2}\hat r_2=\frac{(8.988\times10^9)(14.23\times10^{-9})}{(0.06708)^2}\hat r_2$$ $$\boxed{\vec E_2=(-2.542\times10^4,\,1.271\times10^4,\,0)\ \text{V/m},\ |\vec E_2|=2.842\times10^4\ \text{V/m}}$$ directed at $153.4^\circ$ from $+x$ (second quadrant).
Part (c) — superposition. Adding components: $$\vec E=\vec E_1+\vec E_2=(-3.813\times10^4,\,-0.22,\,0)\ \text{V/m}$$ The $y$-components (12710.59 vs 12710.36 V/m, opposite sign) very nearly cancel — $14.23$ nC is exactly the value that makes this happen — leaving: $$\boxed{\vec E\approx(-3.813\times10^4,\,0,\,0)\ \text{V/m},\ |\vec E|=3.813\times10^4\ \text{V/m, pointing in }-x}$$
Part (d) — force on $Q=-20$ nC placed at $P$. $\vec F=Q\vec E$; the negative charge feels a force opposite to $\vec E$: $$\vec F=(-20\times10^{-9})(-3.813\times10^4,\,0,\,0)$$ $$\boxed{\vec F=(7.626\times10^{-4},\,0,\,0)\ \text{N},\ |\vec F|=7.626\times10^{-4}\ \text{N, pointing in }+x}$$

Quantity	Result
$\vec E_1$	$1.798\times10^4$ V/m at $225^\circ$
$\vec E_2$	$2.842\times10^4$ V/m at $153.4^\circ$
$\vec E$ total	$3.813\times10^4$ V/m, $-x$ direction
$\vec F$ on $Q=-20$ nC	$7.626\times10^{-4}$ N, $+x$ direction

Concept & Theory

Topic. Coulomb's law and the superposition principle for point charges.

Key relations. $\vec E=\dfrac{kq}{r^2}\hat r$, $k=1/(4\pi\varepsilon_0)=8.988\times10^9$ N·m$^2$/C$^2$; field superposition $\vec E_{\text{tot}}=\sum\vec E_i$; force on a charge $\vec F=Q\vec E$.

Why this works. The electric field is linear in source charge, so each point charge's contribution can be computed independently (as a vector, from source to field point) and simply added. A negative test charge experiences a force antiparallel to the local field, not along it.

Common pitfall. Forgetting to normalize the displacement vector to a unit vector before scaling by $kq/r^2$ (i.e. dividing by $r^2$ instead of $r^3$ when working directly with the displacement vector), or dropping the sign of $Q$ when converting field to force.

Source. Sadiku, Elements of Electromagnetics, Ch. 4 (Coulomb's law, electric field of point charges).

Question 1 of 7: Electric Field and Force from Two Point Charges

Question 1: Electric Field and Force from Two Point Charges (20 marks)