Question

Two point charges $q$ and $-q$ are at positions $\left(\right.0, \, 0, \, d\left.\right)$ and $\text{(0, 0, }-d\text{)}$ respectively. What is the electric field at $\left(\right.a, \, 0, \, 0\left.\right)$ ?

• A. $\frac{2 q d}{4 \pi \left(\epsilon \right)_{0} \left(\right. d^{2} + a^{2} \left.\right)^{3 / 2}}\hat{k}$
• B. $\frac{q d}{4 \pi \left(\epsilon \right)_{0} \left(\right. d^{2} + a^{2} \left.\right)^{3 / 2}}\hat{k}$
• C. $\frac{- 2 q d}{4 \pi \left(\epsilon \right)_{0} \left(\right. d^{2} + a^{2} \left.\right)^{3 / 2}}\hat{k}$
• D. $\frac{- q d}{4 \pi \left(\epsilon \right)_{0} \left(\right. d^{2} + a^{2} \left.\right)^{3 / 2}}\hat{k}$

Answer 1

Answer: (C)

At $P$ there are two fields
(1) Due to $A=E_{A}=\frac{k q}{r^{2}}$
(2) Due to $B=E_{B}=\frac{k q}{r^{2}}$
$r=(a^{2}+d^{2})^{1 / 2}$      at angle $\text{180}-\text{2 \theta }$
$E_{n e t}=2E_{A}cos\left[\frac{1}{2} \left(180 - 2 \theta \right)\right]=2E_{A}cos(90-\theta \left.\right)$
$=2E_{A}sin\theta =\frac{2 k q}{r^{2}}\frac{d}{\left(\right. r \left.\right)}$
$=\frac{2 k q d}{\left(\left(\right. a^{2} + d^{2} )\right)^{3 / 2}}$ in $-z$ direction.