Question

Two identical point charges (q = +2.40 x 10^-6 C) are fixed at opposite corners of a square whose sides have a length of 0.350 m. A test charge (q₀ = -5.40 x 10^-8 C), with a mass of 8.80 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

According to the given problem,

Use conservation of energy where (K + U)_i = (K + U)_f

where K = kinetic energy = 1/2*m*v^2

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

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

So V_i = 2*8.99*10⁹*2.40*10^-6/0.350 m = 1.233*10⁵V

and V_f = 2*8.99*10⁹*2.40*10^-6/(√2*0.175) = 1.744*10⁵V

Now we have K_f = 1/2*m*v² = U_i - U_f = -5.40*10^-8*(1.233*10⁵ - 1.744*10⁵)

1/2*m*v² = 2.76x10^-3

so v = √(2*ΔU/m) = 250.43m/s

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