Question

5. A particle of unknown charge obtains 0.042 J of kinetic energy as it moves from point A to point B. Point Ahas an electric potential of 700.0 V; and point B has an electric potential of 200.0 V. Determine the magnitudeand sign of the charge.

Answer 1

Given:

The kinetic energy of the charged particle is,

$\Delta U=0.042\text{ J}$

Point A has electric potential,

$V_A=700.0\text{ V}$

The potential at point B is,

$V_B=200.0\text{ V}$

To find:

The magnitude and the sign of the charge

Explanation:

The energy of the charged particle is,

$q\Delta V$

which is equal to,

$\Delta U$

As the charged particle is moving from higher to lower potential region, the charge is positive. So, we can write,

$\begin{gathered} q(V_A-V_B)=0.042 \\ q(700.0-200.0)=0.042 \\ q=\frac{0.042}{700.0-200.0} \\ q=8.4\times10^{-4}\text{ C} \end{gathered}$

Hence, the charge is positive and the magnitude is,

$8.4\times10^{-4}\text{ C}$