Potential Energy and Work in an Electric Field - Learn

Potential Energy and Work in an Electric Field – Learn

Electric field strength can also be calculated using:

$\vec { E } =\cfrac { V }{ \vec { d } }$

where:

$\vec { E }$ = the electrical field strength (in Vm⁻¹)

$V$ = the electrical potential (in V)

$\vec { d }$ = the distance between points parallel to the field (in m)

The electrical potential, V, is also defined as the potential energy per unit charge at a point and is given by:

$V=\cfrac { \Delta U }{ q }}$

where:

$V$ = the electrical potential (in V or JC⁻¹)

$\Delta U$ = the potential energy (in J)

$q$ = the charge of the point charge (in C)

Calculating work done by an electric field:

Within an electric field, work must be done to move a point charge through the electric field. The source of this work can either be done:

by the electric field on the charged object, or
on the electric field by forcing the object to move

If the charge is moving in the direction that it would naturally be moved by the field then work is being done by the field. If it is moving against the direction it would naturally go then work is done on the field.

To calculate the work done on a point charge to move it through a distance, d, across a potential difference:

$W=qEd$

where:

$W$ = the work done on the point charge or on the field (in J)

$q$ = the charge of the point charge (in C)

$E$ = the electric field strength (in Vm⁻¹ or NC⁻¹)

$d$ = the distance between the points parallel to the field (in m)

Example 1:

What is the strength of an electric field between charged plates separated by a distance of 2cm and a potential of 6V across the plates?

Answer:

using: $\vec { E } =\cfrac { V }{ \vec { d } }$

$\vec { E } =\cfrac { 6 }{ 0.02 }$

$\vec { E } =300\: V{ m }^{ -1 }$

Example 2:

A charged particle of 2.5 × 10⁻¹⁸ C passes through a potential difference of 40V. Calculate the change in potential energy of this particle:

Answer:

using: $V=\cfrac { \Delta U }{ q }}$ and rearranging to give:

$\Delta U=Vq$

$\Delta U=40\times { (2.5\times 10 }^{ -18 })$

$\Delta U={ 1\times 10 }^{ -16 }\:J$

Example 3:

Calculate the work done on an electron that moves 4cm between charged plates with an electric field strength of 20Vm⁻¹: ${ { (q }_{ e }={ -1.602\times 10 }^{ -19 }C) }$

Answer:

using: $W=qEd$

$W=(-1.602\times { 10 }^{ -19 })(20)(.04)$

$W=-1.28\times { 10 }^{ -19 }\:J$