Question

7. Three point-like charges are placed as shown in the figure, r r2 56.0 cm. Find the magnitude of the electric force exerted on the charge q3. Let qi-1.50 C, q23.80 HC, q3-+4.60 HC. q1 43 r1

Is the answer 0.551 N?

Answer 1

electrostatic force is given by

F = kQ1*Q2/d^2

Force will be attractive if both charge have different signs, and force will be repulsive if both charge have same signs.

Force F1, on q3 due to q1 will be attractive and towards the -ve x-axis

Force F2, on q3 due to q2 will be attractive and towards the -ve x-axis

So net force on q3 will be towards -ve x-axis

Fnet = F1 + F2

Fnet = k*q1*q3/d1^2 + k*q2*q3/d2^2

d1 = r1 + r2 = 56.0 + 56.0 = 112 cm = 1.12 m

d2 = r2 = 56.0 cm = 0.56 m

So Using these values:

Fnet = k*q3*[q1/d1^2 + q2/d2^2]

Fnet = 9*10^9*4.60*10^-6*[1.50*10^-6/1.12^2 + 3.80*10^-6/0.56^2]

Fnet = 0.551 N towards -ve x-axis

Magnitude of force = |Fnet| = 0.551 N

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