Answer to Question #171639 in Electric Circuits for Pranali

Question

You create a uniform electric field between two parallel plates; if the electric potential difference between the plates is 125V and the separation of the plates is 0.500m, what final velocity will a pith ball of charge 1.50nC and mass of 0.357g placed next to the positive plate attain?

Answer 1

Explanations & Calculations

• Suppose the plates are placed horizontally & the ball is released near the upper plate which is positive.
• And, if we thought the bottom plate to be the level of zero energy what happens is that the potential energy ($\small mgh$) the ball had at the level of the positive plate & the work done on it ($\small q\Delta V$) as it falls through the gap is converted into kinetic energy at the bottom (negative) plate.
• By the theorem of conservation of energy, the velocity at the bottom plate can be calculated.

$\qquad\qquad \begin{aligned} \small mgh+W&=\small \frac{1}{2}mv^2\\ \small v &=\small \sqrt{\frac{2(mgh+W)}{m}}\\ &=\small\sqrt{\frac{2(0.357\times10^{-3}kg\times9.8\times0.5m+1.5\times10^{-9}C\times125V)}{0.357\times10^{-3}kg}}\\ &=\small\bold{3.131 ms^{-1}} \end{aligned}$