Question

An electron acquires 6.90×10^-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?

$\Delta V_{\rm BA}$ = _________________ V

Which plate is at the higher potential?

Answer 1

Concepts and reason

This problem is based on the concept of work done by an electric field on a charge and work-energy theorem.

The work done is to be estimated by the provided kinetic energy. Then it is to be used to calculate the potential. The plate having higher potential is to be determined.

Fundamentals

Work energy Theorem: The change in the kinetic energy is equal to work done.

Write the expression for the work done on a charge in an electric field.

$W = qV$

Here, $V$ is potential, $q$ is charge and $W$ is work done.

Write the expression for the work-energy theorem.

$W = \Delta K$

Here, $\Delta K$ is change in kinetic energy.

Write the expression for the work-energy theorem.

$W = \Delta K$

Substitute $6.90 \times {10^{ - 16}}\;{\rm{J}}$ for $\Delta K$.

$W = 6.90 \times {10^{ - 16}}\;{\rm{J}}$

Write the expression for the work done on a charge in an electric field.

$W = qV$

$V = \frac{W}{q}$

Substitute $6.90 \times {10^{ - 16}}\;{\rm{J}}$ for $W$ and $1.6 \times {10^{ - 19}}\;{\rm{C}}$ for $q$.

$\begin{array}{c}\\V = \frac{{6.90 \times {{10}^{ - 16}}\;{\rm{J}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C}}}}\\\\ = 4313{\rm{ V}}\\\end{array}$

The electron has negative charge and it is accelerated from plate $A$ to plate $B$.Therefore the plate $B$will have higher potential.

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Ans:

The potential between the plates is $4313{\rm{ V}}$ and plate $B$ has higher potential.