A small sphere of mass m and charge –q is released from rest at point T. If the electric potentials at points S and T are VS and VT, respectively, what is the speed of the sphere when it reaches point S? Ignore the effects of gravity. (A) 2q/m(Vs + VT)
(B) 4q/m(Vs + VT)
(C) q/2m(Vs - VT)
(D) q/2m (Vs + VT)
(E) 2q/m(Vs - VT)

(E) √[2q/m(Vs - VT)]

Explanation:

Since the charge -q moves from VT to VS, the potential difference is VT - VS.

The work done in moving the charge q across a potential difference V is given by W = qV.

Now, the work done in moving the charge -q across that potential difference VT - VS is thus W = -q(VT - VS) = -q[-(VS - VT)] = q(VS - VT). This work equals the sphere's kinetic energy and kinetic energy equals K = 1/2mv² where m = mass of sphere and v = speed of sphere.

So, K = W

1/2mv² = q(VS - VT)

v² = 2q/m(VS - VT)

taking square root of both sides, we have

v = √[2q/m(Vs - VT)]

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