Question

n electron (q=-e) completes half of a circular orbit of radius r around a nucleus with Q=+3e.

a. How much work is done on the electron as it moves from i to f? Give either a numerical value if you knew the radius. Justify your answer.

b. By how much does the electric potential energy change as the electron moves from i to f?

c. Is the electron's speed at f greater than, less than, or equal to its speed at i?

Answer 1

(a) The distance of the electron from the charge \(Q\) is same at all points of its motion. That is from \(i\) to \(f\). So the potential is the same. The work done is, \(W=q \Delta V\)

$$ \begin{array}{l} =q(0) \\ =0 \end{array} $$

(b) The distance of the electron from the charge \(Q\) is the same at all points of its motion. That is from \(i\) to \(f\). So the potential is the same. Hence the electric potential energy is, \(\Delta U=W\)

$$ =0 $$

(c) Apply law of conservation of energy. The change in potential energy is zero. Hence,

$$ \begin{aligned} K_{f} &=K_{i} \\ \frac{1}{2} m v_{f}^{2} &=\frac{1}{2} m v_{i}^{2} \\ v_{f} &=v_{i} \end{aligned} $$

Thus, the final speed is equal to the initial speed.