Electrical Potential Energy

From MyMCAT

Introduction

As briefly discussed in the previous section, charged objects exert electric fields on each other and are capable of pushing or pulling each other with the electrical forces which result. As with all forces, work can be done by or on an object when forces are exerted or applied. Just as a mass a given height off the ground has potential energy due to the gravitational field it is in, so does a charged object in an electric field. In creating a gravitational potential or an electrical potential work must be performed and once the potential has been established it can be converted into energy to perform work. (ie. either the mass falling or a charged particle being pulled to an oppositely charged particle to gain velocity.)

Electrical Potential

Just as work must be performed to combat gravity when raising an object in the air, work may be applied to the concept of electric fields. Consider two positive charges, A and B, separated by a distance r. If B is to be moved closer to A, reducing the distance between them, then work must be performed. Both A and B experience a force repelling them from one another, and thus work must be performed to combat this repulsion and overcome it.

Unlike gravity, however, there are both positive and negative possibilities. Consider, if charge A were negative, then moving B closer to A is actually more favorable, as the charges attract each other anyway, energy would be released, in the form of B gaining velocity towards A. In this case, work would need to be performed when separating the two charges, completely the opposite of the previous case!

Units

As with any other form of energy, electrical potential energy is measured in Volts (V).

Electrical Potential Energy in a Uniform Electric Field

In the case of two parallel and oppositely charged plates, the electric field produced between the two plates is linear and uniform. Any charged particle in between the two plates always feels a push from one plate and a pull from the other, such that no matter where in between the two plates it is, the strength of the force is always the same. (At the edges of the plate or beyond the region which is covered by both plates does not have this property!)

Determining how potential energy is stored in a particle that is between these two plates then can be calculated by how much work would be necessary to move the particle to that position from one side. If we consider a charge q, then the force felt on q must be directly related to the electric field, and the work done to move this charge can be calculated.

$\begin{eqnarray} P.E. = Work &=& Fd \nonumber \\ &=& (qE)d \nonumber \\ &=& qEd \nonumber \end{eqnarray} \nonumber$

Electric Potential Energy Between Two Point Charges

If we consider Coulomb's Law, we can easily determine the force felt by one particle on another. If we then expand this to consider the work required to move one particle from infinity to a specific distance r from another particle we can determine how much work is done. (The derivation requires integration since the force continues to change as the distance changes, but it is not necessary to know on the MCAT, only the final result matters...)

$\begin{eqnarray} P.E. = Work &=& \frac{kq_1q_2}{r} \nonumber \end{eqnarray} \nonumber$

Electrical Potential Energy in combinations of charges

A typical problem is assessing the potential energy in various organizations of particles. With each added particle, the complexity of the problem increases as every particle interacts with every other particle in the system!

One Charged Particle

The electric potential energy of a system containing only one point charge is zero, as no energy is required to move the charge particle to its location in the absence of any electric fields.

Two Charged Particles

The amount of energy to move the first charge to position a is zero. The amount to move the second charge a distance r from the first charge is:

$\begin{eqnarray} P.E. = Work &=& \frac{kq_1q_2}{r} \nonumber \end{eqnarray} \nonumber$

Thus the total is just the one value.

Three Charged Particles

The amount of energy to move the first charge to position a is zero. The amount to move the second charge a distance r from the first charge was given in the previous section. And finally, the amount to move a third charge can be calculated as a sum of moving the charge to the first and second charges individually.

$P.E. = k \left({\frac{q_1 q_2}{r_{12}}} + {\frac{q_1 q_3}{r_{13}}} + {\frac{q_2 q_3}{r_{23}}} \right)$

One can see how this problem expands into a lot of work for larger numbers of charges as each additional charge interacts with all the previous charges...