A charge of 10.8 µC is held fixed at the origin. (a) If a 5.25-µC charge with a

Question

A charge of 10.8 µC is held fixed at the origin.

(a) If a 5.25-µC charge with a mass of 2.03 g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin?

(b) Suppose the 5.25-µC charge is released from rest at the point x = 1/2 (0.925 m) and y = 1/2 (1.17 m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part (a)?

A charge of 10.8 uC is held fixed at the origin. (a) If a -5.25-HC charge with a mass of 2.03 g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin m/s (b) Suppose the -5.25-uC charge is released from rest at the point x 1/2 (0.925 m) and y 1/2 (1.17 m). When it is halfway to the origin, is its speed greater than less than, or equal to the speed found in part (a)? greater than less than equal to Explain Score: 1 out of 1 Comment: (c) Find the speed of the charge for the situation described in part (b). m/s

Explanation / Answer

Given

Charge Q1 = 10.8 x 10-6 C

Charge Q2 = -5.25 x 10-6 C

mass of Q2 is m = 2.03 x 10-3 kg

location of Q1 = (0 m , 0 m )

location of Q2 = (0.925 m 1.17 m)

Solution

Distance between the charges r = [(0.925 – 0)2 + (1.17 – 0)2]1/2

r = 1.49 m

Halfway of the this distance is

s = r/2 = 0.745 m

The force acting on Q2 = kQ1Q2/r2

F = 9 x 109 x 10.8 x 10-6 x 5.25 x 10-6 / 1.492

F = 0.2298 N

Acceleration a = F/m = 0.2298/2.03 x 10-3

a = 113.23 m/s2

using equations of motion

v2 = u2 + 2as

v2 = 02 + 2 x 113.23 x 0.745

v = 12.99 m/s

B)

new location of Q2 = (0.925/2 , 1.17/2) = (0.4625, 0.585)

Distance between the charges r = [(0.4625– 0)2 + (0.585– 0)2]1/2

r = 0.746 m

Halfway of the this distance is

s = r/2 = 0.373 m

The force acting on Q2 = kQ1Q2/r2

F = 9 x 109 x 10.8 x 10-6 x 5.25 x 10-6 / 0.7462

F = 0.9169 N

Acceleration a = F/m = 0.9169 /2.03 x 10-3

a = 451.70 m/s2

using equations of motion

v2 = u2 + 2as

v2 = 02 + 2 x 451.70 x 0.373

v = 18.36 m/s

This speed is greater than the speed of first scenario.

This happens because in the second scenario the force acting on Q2 is higher so the acceleration is also higher. Hence the change in velocity increases even more gradually

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