Atest charge of +2μc is placed halfway between a charge of +6μc and another of +4μc separated by 10 cm. (a) what is the magnitude of the force on the test charge? (b) what is the direction of this force (away from or toward the +6μc charge)?

A) The magnitude of the force on the test charge is; F = 14.384 N

B) The direction of the net force will be; away from the +6μC charge.

We are given;

q = +2μC = 2 × 10^(-6) C

q1 = +6μC = 6 × 10^(-6) C

q2 = +4μC = 4 × 10^(-6) C

r = 10/2 = 5 cm = 0.05 m (divided by 2 because the charge is placed halfway)

A) Magnitude of the force on the test charge is given by the formula;

F = (Kq1*q/r²) - (Kq2*q/r²)

Where k = 8.99 × 10^(9) N.m²/C²

Thus;

F = ((8.99 × 10^(9) × 6 × 10^(-6) × 2 × 10^(-6))/(0.05²)) - ((8.99 × 10^(9) × 4 × 10^(-6) × 2 × 10^(-6))/(0.05²))

F = 14.384 N

B) The charge of +6μC is greater than that of +4 μC in magnitude and therefore, the net force will be in a direction away from the +6μC charge.

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Part a)

F = 14.4 N

Part b)

away from +6μC charge

Explanation:

Force on the charge placed midway between two charges is given by

here we know that

here the charge is placed at mid point so we have

now we have

Part b)

since +6μC charge is more in magnitude so the force due to this charge will be more so the net force on it is away from +6μC charge

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