Question

A test charge of 10-6C is placed halfway between a charge of 5·10-6C and a
charge of 3·10-6C that are 20 cm apart. Find the magnitude and direction of the force
on the test charge.

Answer 1

Answer:

EC

Explanation:

Answer 2

$F=1.8\ N$ is the net force that acts on the test charge in a direction away from the second charge.

Explanation:

given:

• quantity of test charge, $q=10^{-6}C$

• magnitude of first charge, $q_1=5\times 10^{-6}C$

• magnitude of second charge, $q_2=3\times 10^{-6}\ C$

• distance between the two charges, $r=20\ cm=0.2\ m$

• Distance of test charge from the first charge, $d_1=10\ cm=0.1\ m$

• Distance of test charge from the second charge, $d_2=10\ cm=0.1\ m$

Force on the test charge due to first charge:

$F_1=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q}{d_1^2}$

$F_1=9\times 10^9\times \frac{5\times 10^{-6}\times 10^{-6}}{0.1^2}$

$F_1=4.5\ N$

Force on the test charge due to second charge:

$F_2=\frac{1}{4\pi.\epsilon_0}\times \frac{q_2.q}{d_2^2}$

$F_2=9\times 10^9\times \frac{3\times 10^{-6}\times 10^{-6}}{0.1^2}$

$F_2=2.7\ N$

Since the test charge is at the midway of the two charges, therefore the direction of the force will be away from the charge of greater charge and the magnitude of the force will be the difference of the forces due to two charges.

$F=F_1-F_2$

$F=4.5-2.7$

$F=1.8\ N\ ( towards\ q_1)$

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TOPIC: electrostatic force

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