Question

A stationary proton is moved from point A, where the potential is 450 V, to point B, where the potential is 125 V.

a.) How much work is done by the electric force?

b.) What is its speed at point B?

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Answer 1

a) By the definition of the work done, we get:

$$\Delta V=V_B-V_A=-\dfrac{W}{q},$$ $$W=-q(V_B-V_A),$$ $$W=-1.6\cdot10^{-19}\ C\cdot(125\ V-450\ V)=5.2\cdot10^{-17}\ J.$$

(b) Let's apply the law of conservation of energy:

$$KE=PE,$$ $$\dfrac{1}{2}mv^2=qV,$$ $$v=\sqrt{\dfrac{2qV}{m}}=\sqrt{\dfrac{2\cdot1.6\cdot10^{-19}\ C\cdot125\ V}{1.67\cdot10^{-27}\ kg}}=1.55\cdot10^{5}\ \dfrac{m}{s}.$$