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# Potential Energy Assignment Help

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Electromagnetism - Potential Energy

**Potential Energy**

It is the amount of work done to bring a charge q from infinity to that point against the electric fields of a given charge without changing its kinetic energy.

**PE = qQ/4πε**_{0}r = qV

Work done **W = ?PE = PE **_{final} - PE _{initial}

**= q**_{1}q_{2}/4πε_{0} [(1/r_{f}) - (1/r_{1}]

Force on a surface charge:- the repulsive force acting on an element of a charged surface due to rest of the charged surface is called electric force on a charged conducting surface.

**E = E**_{!} + E_{2} = σ/ε_{0} where **E**_{1} is the electric field due to small element and **E**_{2} is due to rest of the charge.

**E**_{1} = E_{2}

Therefore electric field intensity near a charged surface =** σ/ 2ε**_{0} and force **dF = σ**^{2} / 2ε_{0 } ds

Therefore **F = ?σ**^{2} / 2ε_{0} ds

And pressure **p **_{elect.} = dF / ds = σ^{2} / 2ε_{0} = ε_{0} E_{2} / 2

In case of a soap bubble

P _{in} - p _{out} = P _{excess} = P _{ST} = P_{ elect.} = 4T / r = σ^{2} / 2ε_{0}

**= 4T / r - q**^{2} / 2 A^{2} ε^{2}_{0} = 4T / r - q^{2} / 32 π^{2} r_{4} ε_{0}

If air pressure inside and outside the bubble are equal then

**P**_{in} = P_{out} or 4T / r = q^{2} / 32 π^{2} r_{4} ε_{0}

Energy density = energy per unit volume

U = ε_{0} E ^{2} / 2 = σ^{2} / 2ε_{0}

Total energy **U**_{ tot} = ∫ε_{0} E2 / 2 dV

Charged liquid drop if a identical drops each of radius r and charge q are joined to form a big drop of radius R and charge Q

Then **R = n**_{1}/3 r Q big = nq small

**E **_{big} = n_{1}/3 E _{small}; V_{big} = n_{2}/3 V _{small}; σ _{big} = n1/3 σ _{small}

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**Potential Energy**

**PE = qQ/4πε**

_{0}r = qVWork done

**W = ?PE = PE**

_{final}- PE_{initial}

**= q**

_{1}q_{2}/4πε_{0}[(1/r_{f}) - (1/r_{1}]Force on a surface charge:- the repulsive force acting on an element of a charged surface due to rest of the charged surface is called electric force on a charged conducting surface.

**E = E**

_{!}+ E_{2}= σ/ε_{0}where

**E**

_{1}is the electric field due to small element and

**E**

_{2}is due to rest of the charge.

**E**

_{1}= E_{2}

Therefore electric field intensity near a charged surface =

**σ/ 2ε**and force

_{0}**dF = σ**

^{2}/ 2ε_{0 }ds

Therefore

**F = ?σ**

^{2}/ 2ε_{0}dsAnd pressure

**p**

_{elect.}= dF / ds = σ^{2}/ 2ε_{0}= ε_{0}E_{2}/ 2In case of a soap bubble

P

P

_{in}- p_{out}= P_{excess}= P_{ST}= P_{ elect.}= 4T / r = σ^{2}/ 2ε_{0}

**= 4T / r - q**

^{2}/ 2 A^{2}ε^{2}_{0}= 4T / r - q^{2}/ 32 π^{2}r_{4}ε_{0}

If air pressure inside and outside the bubble are equal then

**P**

_{in}= P_{out}or 4T / r = q^{2}/ 32 π^{2}r_{4}ε_{0}

Energy density = energy per unit volume

U = ε

U = ε

_{0}E^{2}/ 2 = σ^{2}/ 2ε_{0}

Total energy

**U**

_{ tot}= ∫ε_{0}E2 / 2 dVCharged liquid drop if a identical drops each of radius r and charge q are joined to form a big drop of radius R and charge Q

Then

**R = n**small

_{1}/3 r Q big = nq**E**

_{big}= n_{1}/3 E_{small}; V_{big}= n_{2}/3 V_{small}; σ_{big}= n1/3 σ_{small}

ExpertsMind.com - Potential Energy Assignment Help, Potential Energy Homework Help, Potential Energy Assignment Tutors, Potential Energy Solutions, Potential Energy Answers, Electromagnetism Assignment Tutors