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

Answer:

Given :-

First Charge = {q}_{1}q

1

= 6μC = 6×10-⁶ C

Second Charge = {q}_{2}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}F=14.4N

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

Explanation:

hope it will help you a lot

Answer 2

Consider q1 = 6 uC = 6 x 10^-6 C

q2 = 4 uC = 4 x 10^-6 C

and q = 2 uC = 2 x 10^-6 C

r = 10/2 = 5 cm = 0.05 m

(a) Force on q due to charge q1 -

F1 =...