Question

45. Predict/Calculate Figure 19-42 shows a system consisting of three charges, q1 +5.00 uC, 92 +5.00 uC, and gs5.00 uC, at the vertices of an equilateral triangle of side d 2.95 cm. (a) Find the magnitude of the electric field at a point halfway between the charges qi and q2. (b) Is the magnitude of the electric field halfway between the charges q2 and qs greater than, less than, or the same as the electric field found in part (a)? Explain. (c) Find the magnitude of the electric field at the point specified in part (b)

Answer 1

Force due to Electric field at a point = KQ/r2

Assume a unit positive charge placed at a point midway between q1 and q2. Then F is due to q1, q2 and q3.

It is easy to see that the line joining q1 and q2 (take as X axis) is perpendicular to the line joining the midpoint and q3 (take as Y axis).

F1 = Kq1/r12 = 9*10^9 * 5 * 10^-6 / ((2.95 * 10^-2)/2)2 = 206.84 * 10^6 N/C (repulsive) along +ve X

F2 = Kq2/r22 = 9*10^9 * 5 * 10^-6 / ((2.95 * 10^-2)/2)2 = 206.84 * 10^6 N/C (repulsive) along -ve X

F3 = Kq3/r32

r3 = altitude of the equilateral triangle = √3/2*d = 0.0255 m

= 9*10^9 * -5 * 10^-6 / (0.0255)^2

∴ F3 = 69.204 * 10^6 N/C (attractive) along +ve Y

Since q1 and q2 lie along a straight line the force along that line is simply F1 + F2 = 0

Also, there will be a force due to q3 in the perpendicular direction, F3

a.) Magnitude of electric field = ((F1 + F2)2 + F32) ^ 1/2 = F3 = 69.204 * 10^6 N/C

b.) In this case, since q2 and q3 lie along the same line q2 repels and q3 attracts and it's easy to see that both act in the same direction. Hence the net force is F2 + F3 which is not zero. Additionally, there is also the force due to q1 whose magnitude will remain the same as that due to q3 in the previous case. Therefore the force in this case will be greater than in the previous case.

c.)

It is easy to see that the line joining q2 and q3 (take as X axis) is perpendicular to the line joining the midpoint and q1 (take as Y axis).

F1 = 69.204 * 10^6 N/C (repulsive)

     F2 = 206.84 * 10^6 N/C (repulsive)

     F3 = 206.84 * 10^6 N/C (attractive)

     ∴ Electric field is ((F1 + F2)2 + F32) ^ 1/2 = 419.43 * 10^6 N/C