Question

The electric potential along the x-axis is V=100e?2x/mV, where x is in meters.

Part A

What is Ex at x=1.0m?

Express your answer as an integer and include the appropriate units.

ANSWER:

Ex =

Part B

What is Ex at x=2.4m?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Ex =

Answer 1

Concepts and reason

The main concept used to solve the problem is relation between electric field and potential difference.

Initially, use the expression to calculate the electric potential to calculate the expression of electric field. Later, use this expression to calculate the electric field at $x = 1.0\,{\rm{m}}$ . Finally, use the expression of electric field to calculate the electric field at $x = 2.4\,{\rm{m}}$ .

Fundamentals

The relation between electric field and electric potential is expressed as follows:

$E = - \frac{{dV}}{{dx}}$

Here, E is the electric field, dV is the change in potential, and dx is the change in the distance.

(A)

Calculate the electric field at $x = 1.0\,{\rm{m}}$ .

The relation between electric field and electric potential is expressed as follows:

$E = - \frac{{dV}}{{dx}}$

Here, E is the electric field, dV is the change in potential, and dx is the change in the distance.

Substitute $100{e^{ - 2x}}\,{\rm{V}}$ for V in expression $E = - \frac{{dV}}{{dx}}$ .

$\begin{array}{c}\\{E_x} = - \frac{d}{{dx}}\left( {100{e^{ - 2x}}\,{\rm{V}}} \right)\\\\ = - 100\frac{d}{{dx}}\left( {{e^{ - 2x}}\,{\rm{V}}} \right)\\\\ = 200{e^{ - 2x}}\\\end{array}$

Now, Substitute 1.0 m for x in expression ${E_x} = 200{e^{ - 2x}}$ .

$\begin{array}{c}\\{E_x} = 200{e^{ - 2\left( {1.0\,{\rm{m}}} \right)}}\\\\ = 27\,{\rm{V/m}}\\\end{array}$

(B)

Calculate the electric field at $x = 2.4\,{\rm{m}}$ .

The relation between electric field and electric potential is expressed as follows:

$E = - \frac{{dV}}{{dx}}$

Here, E is the electric field, dV is the change in potential, and dx is the change in the distance.

Substitute $100{e^{ - 2x}}\,{\rm{V}}$ for V in expression $E = - \frac{{dV}}{{dx}}$ .

$\begin{array}{c}\\{E_x} = - \frac{d}{{dx}}\left( {100{e^{ - 2x}}\,{\rm{V}}} \right)\\\\ = - 100\frac{d}{{dx}}\left( {{e^{ - 2x}}\,{\rm{V}}} \right)\\\\ = 200{e^{ - 2x}}\\\end{array}$

Now, Substitute 2.4 m for x in expression ${E_x} = 200{e^{ - 2x}}$ .

$\begin{array}{c}\\{E_x} = 200{e^{ - 2\left( {2.4\,{\rm{m}}} \right)}}\\\\ = 1.64\,{\rm{V/m}}\\\end{array}$

Ans: Part A

The magnitude of electric field at point $x = 1.0\,{\rm{m}}$ is $27{\rm{ V/m}}$ .

Part B

The magnitude of electric field at point $x = 2.4\,{\rm{m}}$ is $1.64{\rm{ V/m}}$ .