1)A particle of mass 0.002 gram and carrying a charge of 2 coulombs moves from r

Question

1)A particle of mass 0.002 gram and carrying a charge of 2 coulombs moves from rest through a region of potential difference of 6 V. What is the final speed of the particle?

2) Three point charges are placed along the x-axis. A 50 nC charge is placed at x = 1.0 m, another charge of -100 nC is placed at x = -0.3 m. The third charge of 70 nC is placed at x = -0.7 m. Assume the electric potential at infinity is zero. Calculate the electric potential at the origin.

3) A point charge, 12 C, is placed at the origin while another point charge, 6.5 C, is placed at the position y = 5.0 cm. What is the potential energy of this arrangement?

Explanation / Answer

1. delta(KE) = q deltaV

(0.002 x 10^3 kg) v^2 / 2 =(2 C) (6 V)

v = 3464 m/s

2. V = k q1 / r1 + k q2 / r2 + k q3 / r3

V = (9 x 10^9)[(50 x 10^-9/1) + (-100 x 10^-9 / 0.3) + (70 x 10^-9/0.7)]

V = - 1650 Volt

3. PE = k q1 q2 / r

Pe = (9 x 10^9)(12 x 10^-6) (6.5 x 10^-6) / 0.05

= 14.04 J

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