Question

a test charge of 2uc is placed halfway between a charge of 6 uc and 4uc separated by 10 cm . find the magnitude of the force on the test charge

Answer 1

Given :-

First Charge = ${q}_{1}$= 6μC = 6×10-⁶ C

Second Charge = ${q}_{2}$= 4μC = 4 × 10-⁶ C

Mid point test charge = q = 2μC = 2×10-⁶ C

Distance between them = 5 cm = 5 × 10-²m

There will be two forces acting on the test one is due to q1 and other is due to q2. Hence the net charge acting on test charge is F.

F = F(due to q1) + F(due to q2)

As both the forces are acting in opposite.

F = kq1q/r² - kq2q/r²

F = k/r²[6 × 10-⁶ × 2 × 10-⁶ - 4 ×10-⁶ × 2 ×10-⁶]

F = 9×10⁹/25 × 10-⁴ [12 - 8] × 10-¹²

F = (9 × 4 × 10)/25

F = 360/25

$\bold{F = 14.4\: N}$

Hence, the force acting on the test charge = F = 14.4 N

Answer 2

Given :-

A  test charge of 2uc is placed halfway between a charge of 6 uc and 4uc separated by 10 cm

To Find :-

Magnitude

Solution :-

We know that

$\sf F = \dfrac{kq_1q}{r^2} - \dfrac{kq_2q}{r^2}$

$\sf F = \dfrac{k}{r^2 \times q_1 \times q_2 \times q}$

$\sf F = \dfrac{9 \times 10^9}{25\times6 \times 10^{-6}\times 4 \times 10^{-6} \times 2 \times 10^{-6}}$

$\sf F = \dfrac{9\times 10^9}{25 \times 10^{-4} \times 4\times 10^{-12}}$

$\sf F = \dfrac{9\times 10^9}{100 \times 10^{-8}}$

$\sf F = 14.4 \; N$