Question

Calculate the total electric field $\overrightarrow{E}$ at position A, at a distance $a$ from the y-axis.

• A. $\frac{1}{4\pi {ϵ}_0}\frac{2qd}{(a^2+d^2{)}^{3/2}}\hat{i}$
• B. $\frac{1}{4\pi {ϵ}_0}\frac{qd}{(a^2+d^2{)}^{3/2}}\hat{j}$
• C. $\frac{1}{4\pi {ϵ}_0}\frac{qd}{(a^2+d^2{)}^{3/2}}\hat{i}$
• D. $\frac{1}{4\pi {ϵ}_0}\frac{2qd}{(a^2+d^2{)}^{3/2}}\hat{j}$

Answer 1

The correct option is D $\frac{1}{4\pi {ϵ}_0}\frac{2qd}{(a^2+d^2{)}^{3/2}}\hat{j}$

E.field at A due to $-q$

$E_q=\frac{K(-q)}{(d^2+a^2)}$

E.field due to $+q$ at A

$E_{+q}=\frac{K(+q)}{(d^2+a^2)}$

Resulting E-field at A due to $+q$ & $-q$ will be

$E_r=E_{-q}\cos \theta +E_{+q}\cos \theta$

$=\frac{Kq}{(d^2+a^2)}\times \frac{d}{{(d^2+a^2)}^{1/2}}+\frac{Kqd}{(d^2+a^2){(d^2+a^2)}^{1/2}}$

$=\frac{2Kqd}{{(d^2+a^2)}^{3/2}}\hat{J}$

$\{E_A=\frac{2qd}{4\pi {\epsilon }_0{(a^2+d^2)}^{3/2}}\hat{J}\}$

Hence Option (D) is correct.