Question

Two identical point charges (q = +6.70 x 10-6 C) are fixed at opposite corners of a square whose sides have a length of 0.550 m. A test charge (q0 = -7.80 x 10-8 C), with a mass of 3.20 x 10-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square.

Answer 1

Use conservation of energy where (K + U)1 = (K + U)2 K = kinetic energy = 1/2*m*v^2

and U is potential energy = V*q where V is the potential

Initially K1 = 0 so we need the V at points 1 and 2 (Recall V = k*Σq/r)

So V1 = 2*9.0x10^9*6.70x10^-6/0.550 m = 2.192*10^5V

and V2 = 2*k*6.70x10^-6/(sqrt(2)*0.275) = 1.6261x10^5V

Now we have K2 = 1/2*m*v^2 = U1 - U2 = - 7.80x10^-8*(2.192*10^5 - 1.626x10^5) = -4.4148x10^-3

so v = sqrt(2*ΔU/m) = sqrt(2*4.4148x10^-3/3.20x10^-8) = 525.285m/s