Question

2. Consider a line of charges q1 and q3 = 10.0 μ( at 4.0 cm from the origin, what is the force on q3 due to the other two charges? 80pC at the origin, q,--12.0 μC at 2.0 cm from the origin, -0 เป็2M

Answer 1

Electrostatic force is given by:

F = k*Q1*Q2/R^2

Where force will be attractive if both charge have opposite signs and force will be repulsive if both charges have same sign

Now net force on charge q3 is given by

F13 = Force on q3 due to q1, force will be repulsive towards the +ve x-axis

F23 = Force on q3 due to q2, force will be attractive towards the -ve x-axis

So,

Fnet = F13 - F23

Fnet = k*q1*q3/r13^2 - k*q2*q3/r23^2

r13 = distance between q1 and q3 = 4 cm = 0.04 m

r23 = distance between q2 and q3 = 2 cm = 0.02 m

q1 = 8*10^-6 C, q2 = -12*10^-6 C, q3 = 10*10^-6 C

k = electrostatic constant = 9*10^9

So,

Fnet = 9*10^9*8*10^-6*10*10^-6/0.04^2 - 9*10^9*12*10^-6*10*10^-6/0.02^2

Fnet = -2250 N

(-ve sign means net force on q3 will be towards left side (or -ve x-axis))

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