1. Two fixed point charges are separated with a distance of 6 cm (their coordina

Question

1. Two fixed point charges are separated with a distance of 6 cm (their coordinates are shown in the figure) and each one carries the charge of +Q=3uC. a) Calculate the electric potential at points A( x-axis) and B(y-axis), whose coordinates are A(0 cm, 0 cm) and B (0 cm, 4 cm) A test charge of mass m= 0.001 g and charge –q=-.1 uC is placed at point B, and then released from rest. Will this test charge pass point A? If yes, calculate the speed of the test charge when it passes point A.(Ignore the gravity)

Explanation / Answer

Here ,

at the point A

electric potential is VA

VA = 9 *10^9 * 2 * 3 *10^-6/0.03

VA = 1.8 *10^6 V

Now , at point B

VB = 9*10^9 * 2 * 3 *10^-6/sqrt(0.03^2 + 0.04^2)

VB = 9*10^9 * 2 * 3 *10^-6/0.05

VB = 1.08 *10^6 V

as the electric field is in the direction of decreasing electric potential

electric field is from A to B

the force on the -q will be from B to A

hence , the particle will pass through point A

Now , let the speed is v

0.5 * 1 *10^-6 * v^2 = 0.1 * 10^-6 *(1.8 *10^6 - 1.08 *10^6)

v = 379.5 m/s

the speed of test charge is 379.5 m/s

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