Question

Four identical particles (charge Q=1.6 nC, mass m = 4 g) are placed at corners of a square of side b = 5.9 cm. What is the electric potential energy of this system? (The electric potential energy is zero when the particles are very far apart.)

The particles are now released from rest. What is the speed of each particle when they are very far apart? (You may ignore gravity.)

Answer #1

Here ,

q = 1.6 nC

electric potential energy of the system = total energy between all the charges

electric potential energy of the system = 4 * k * q^2/b + 2 * k * q^2/(sqrt(2) * b)

electric potential energy of the system = 9 *10^9 * (1.6 *10^-9)^2 * (4/.059 + 2/(sqrt(2) * 0.059)

electric potential energy of the system = 2.11 *10^-6 J

the electric potential energy of the system = 2.11 *10^-6 J

----------------------

let the velocity of each at very far is v

decrease in potential energy = increase in kinetic energy

4 * 0.50 * m * v^2 = 2.11 *10^-6

2 * 0.004 * v^2 = 2.11 *10^-6

solving for v

**v = 0.0162 m/s**

**the speed of each very far away is 0.0162
m/s**

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