Question

An electron acquires 5.25 X 10^-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential?

Answer 1

Concepts and reason

The concept required to solve the given problem is relation between potential difference and kinetic energy.

Initially, find the expression for the potential difference between the plates with the help of the relation between potential difference and kinetic energy.

Substitute the given values in the expression to calculate the value of potential difference.

Fundamentals

The relation between the kinetic energy acquired by charged particles and the potential difference between the plates where the electric field is applied is given by,

$e\Delta V = K$

Here, $\Delta V$ is the potential difference between the plates, $e$ is the charge on an electron and $K$ is the kinetic energy.

The kinetic energy acquired by the electron when it is accelerated by an electric field is,

$K = 5.25 \times {10^{ - 16}}{\rm{ J}}$

The kinetic energy acquired by the electron will be proportional to the potential difference between the plates across which the electric field is applied. It will be given by,

$e\Delta V = K$

Here, $\Delta V$ is the potential difference between the plates, $e$ is the charge on an electron and $K$ is the kinetic energy.

Rearrange the above expression.

$\Delta V = \frac{K}{e}$

The potential difference between the plates will be,

$\Delta V = \frac{K}{e}$

Substitute$5.25 \times {10^{ - 16}}{\rm{ J}}$for $K$ and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $e$in the above equation.

$\begin{array}{c}\\\Delta V = \frac{{5.25 \times {{10}^{ - 16}}{\rm{ J}}}}{{1.60 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\ = 3.28 \times {10^3}{\rm{ V}}\left( {\frac{{1{\rm{ kV}}}}{{1000{\rm{ V}}}}} \right)\\\\ = 3.28{\rm{ kV}}\\\end{array}$

Since the electrons will be accelerated towards the plate at higher potential hence plate B will be at higher potential.

Ans:

The potential difference between the plates is $3.28{\rm{ kV}}$and plate B is at higher potential.