Question

Two fixed charges, A and B, are located at x axis. A is at (-3 m, 0 m), B is at x = (+3 m, 0 m). QA = +4.0 μC and QB = -4.0 μC. Calculate the electric field at point (0 m, 4 m) on the y axis.

Answer 1

Answer:$2.88\times 10^3$

Explanation:

Given

Charge on A is $4 \mu C$ at (-3,0)

Charge on B is $-4 \mu C$ at (3,0)

Electric Field at a distance d is given by $E=\frac{kQ}{d^2}$

Thus Electric filed at (0,4) due to A is

$E_A=\frac{kQ_A}{d^2}=\frac{1}{4\pi \epsilon }\frac{Q_A}{d^2}$

Here $d =\sqrt{3^2+4^2}=5$

$E_A=\frac{9\times 10^9\times 4\times 10^{-6}}{5^2}$

$E_A=1.44\times 10^3 N\C (Away\ From\ Q_A)$

Similarly $E_B$ is $1.44\times 10^3 N\C (towards Q_B)$

Electric Field is at an angle $\theta$which is given by

$tan\theta =\frac{4}{3}$

Thus $E_Asin\theta$and $E_Bcos\theta$component will cancel out each other and Cos component will add up

$E_{net}=E_Acos\theta +E_Bcos\theta$

$E_{net}=2Ecos\theta$

$E_{net}=2\times 1.44\times 10^3 N\C$   (towards positive x axis )