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# Potential Energy In External Field Assignment Help

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Electrostatics - Potential Energy In External Field

**Potential Energy in External Field**

**~ - Represents Vector**

Potential energy of a single charge in an external field

We have already obtained an expression for potential energy of a system of charges. In this case, the source of the electric field, i.e. charges and their locations – was specified.

We have now to determine potential energy of a charge (or charges) in an external field ~E - which is not produced by the given charges whose potential energy we have to calculate. The sources which produce external field ~E are often unknown and they are of no interest to us.

The external electric field ~E and the corresponding external potential V may change from point to point. If**V (~r)** is extended potential at any point P of position vector ~r, then by definition, work done in bringing a unit positive charge from infinity to the point P is equal to V.

Therefore, work done is bringing a charge q from infinity to the point P in the external field **= q . V (~r).**

This work is stored in the charged particle in the form of its potential energy.

Potential energy of a single charge q at ~r in an external field = q . V (~r).

Potential energy of a system of two charges in an external field

Suppose **q**_{1}, q_{2} are two mid point charges at position vectors ~r_{1} and ~r_{2}_{ }respectively, in a uniform external electric field of intensity ~E To calculate the potential energy of the system of these two charges in the external field, we find that

Work done in bringing charge **q**_{1} from infinity to position ~r_{1} **is W**_{1} = q_{1} . V (~r_{1} **) **where V (~r_{1} ) is potential at ~r_{1} due to external field.

Again, work done in bringing charge **q**_{2} from infinity to position ~r_{2} against the external field is **W**_{2} = q_{2} . V (~r_{2}) where V (**~r**_{2}) is potential at **~r**_{2} due to external field.

While bringing** q**_{2} from infinity to position **~r**_{2} , work has also to be done against the field due to **q**_{1}. This is

**W**_{3} = q_{1}q_{2}/4∏?_{0} **~r**_{1}_{2}

Where **~r**_{1}_{2} is the distance between **q**_{1} and **q**_{2}.

By the superposition principle, we add up the work done and find P.E. of the system = total work done in assembling the charge configuration

**U = q**_{1} . V (**~r**_{1}**) + q**_{2} . V (**~r**_{2} ) + q_{1}q_{2}/4∏?_{0} **~r**_{1}_{2}

Potential Energy In External Field Assignment Help, Potential Energy In External Field Homework Help, Potential Energy In External Field Tutors, Potential Energy In External Field Solutions, Potential Energy In External Field Tutors, Electrostatics Help, Physics Tutors, Potential Energy In External Field Questions Answers

**Potential Energy in External Field**

**~ - Represents Vector**

We have already obtained an expression for potential energy of a system of charges. In this case, the source of the electric field, i.e. charges and their locations – was specified.

We have now to determine potential energy of a charge (or charges) in an external field ~E - which is not produced by the given charges whose potential energy we have to calculate. The sources which produce external field ~E are often unknown and they are of no interest to us.

The external electric field ~E and the corresponding external potential V may change from point to point. If

**V (~r)**is extended potential at any point P of position vector ~r, then by definition, work done in bringing a unit positive charge from infinity to the point P is equal to V.

Therefore, work done is bringing a charge q from infinity to the point P in the external field

**= q . V (~r).**

This work is stored in the charged particle in the form of its potential energy.

Potential energy of a single charge q at ~r in an external field = q . V (~r).

Potential energy of a system of two charges in an external field

Suppose

**q**

_{1}, q_{2}are two mid point charges at position vectors ~r

_{1}and ~r

_{2}

_{ }respectively, in a uniform external electric field of intensity ~E To calculate the potential energy of the system of these two charges in the external field, we find that

Work done in bringing charge

**q**

_{1}from infinity to position ~r

_{1}

**is W**~r

_{1}= q_{1}. V (_{1}

**)**where V (~r

_{1}) is potential at ~r

_{1}due to external field.

Again, work done in bringing charge

**q**

_{2}from infinity to position ~r

_{2}against the external field is

**W**where V (

_{2}= q_{2}. V (~r_{2})**~r**) is potential at

_{2}**~r**due to external field.

_{2}While bringing

**q**from infinity to position

_{2}**~r**, work has also to be done against the field due to

_{2}**q**

_{1}. This is

**W**

_{3}= q_{1}q_{2}/4∏?_{0}**~r**

_{1}_{2}Where

**~r**is the distance between

_{1}_{2}**q**

_{1}and

**q**

_{2}.

By the superposition principle, we add up the work done and find P.E. of the system = total work done in assembling the charge configuration

**U = q**

_{1}. V (**~r**

_{1}**) + q**

_{2}. V (**~r**) + q_{2}_{1}q_{2}/4∏?_{0}**~r**

_{1}_{2}