Two particles, with charges of 20.0 nC and -20.0 nC, are placed at points with c

Question

Two particles, with charges of 20.0 nC and -20.0 nC, are placed at points with coordinates (0, 4.00 cm) and (0, -4.00 cm) as shown in the figure. A particle with charge 10.0 nC is located at the origin.
(a) Find the electric potential energy of the configuration of the three fixed charges.

(b) A fourth particle, with a mass of 2.00 x 10^-13 kg and charge of 40.0 nC, is released from rest at the point (3.00cm, 0). Find its speed after it has moved freely to a very large distance away.

Explanation / Answer

A) Potential energy is U = (9*10^9)*((-20*20/0.08)+(20*10/0.04)+(-20*10/0.04)) = 4.5*10^-5 J

B) let V be the potentia at 40nC

then V = (9*10^9)*((20/0.05)-(20/0.05)+(10/0.03))*10^-9 = 3000 volts

at infinity potential is 0 V

then apply the relpation W = 0.5*m*v^2

but W = q*V = 0.5*m*v^2

speed is v = sqrt(q*V/(0.5*m))= sqrt(40*10^-9*3000/(0.5*2*10^-13)) = 3.46*10^4 m/s

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