In Figure, a battery of potential difference V=12 Vis connected to a resistive strip of resistance $\mathit{R}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\Omega }$.When an electron moves through the strip from one end to the other, (a) In which direction in the figure does the electron move,(b) How much work is done on the electron by the electric field in the strip, and (c) How much energy is transferred to the thermal energy of the strip by the electron?

1. The potential difference connected to the strip, V=12V
2. The resistance of the strip,$R=60\Omega$

From the sign of the battery, we can predict the direction of the electron motion. Since work done on the electron by the electric field is equal to the change in potential energy of the electron, we can find it using a corresponding formula. This energy is converted into the thermal energy of the strip.

Formula:

The potential energy due to the potential difference in the system, U=qV …(i)

The direction of motion of the electron will be from the negative terminal to the positive terminal, i.e. from lower potential to the higher potential. This is because the electron is a negatively charged particle.

Hence, the direction of the electrons in the strip will be upward.

The work done by the electric field on the electron will be equal to the change in potential energy of electron.Thus, the work done on the electron by the fieldcan be calculated using equation (i) as follows:

$\begin{array}{l}U=1e\times 12\\ =12ey\end{array}$

Hence, the value of the work done is 12 eV.

The potential energy change for the electron will be released as a thermal energy to the strip as there is no energy dissipation.

Hence, the thermal energy transferred will also be 12eV.