Electrical Charge

Introduction

Outline

• Electric Potential
  • Electrical Potential Energy
  • Electrical Potential Energy and Electrical Fields
  • Dipoles
• Practice with the Concepts
• Discussion Points

Electric Potential

We now turn from looking at the results of putting point charges in the fields surrounding a primary charge (the test charge moves), to looking at the attributes of the field itself.

Electrical Potential Energy

We have defined potential energy as the energy stored in a system, and we have always measured potential energy as the difference between two states of the system, never as some absolute amount (as we do with kinetic energy). The potential energy of a rock on top of a mountain is always compared to its potential energy at the bottom of the mountain, or halfway between, or someplace out in space—there is no zero point for potential energy. We calculate the difference by determining the amount of work involved in moving an object from one potential energy state to another.

In the same manner, we will talk about electrical potential energy as the difference in the force experienced by a test charge placed in a non-uniform electrical field. We determine electrical potential

• as a function of the change in potential energy the charge "stores" at each point, or
• as a function of the work necessary to move the charge from one point to the other:

$$V_a=\frac{{\mathrm{PE}}_a}{q}$$

The unit of electrical potential is the Volt, after Volta, who created the first stable electrical batteries. Because of this unit name, electric potential is also referred to as voltage. If we say there is a voltage of 50 Volts, we mean there is a difference of 50 volts between some zero point or ground and another point. We can never measure the absolute potential of the field, only the difference in potential between different points.

Electrical Potential Energy and Electrical Fields

We have already established relationships between work and force W = F * d, and between electrical fields and electrical force: F_E = qE. We can now manipulate these values to get some more relationships:

$$W=\mathrm{Fd}=\mathrm{qEd}$$

Therefore

$$V=\mathrm{Ed}$$ and $$E=V/d$$

This last is very useful: it means the field is equal to the difference in potential between two points divided by the distance that separates them. Because work is measured in units of energy, we can talk about the work necessary to move charges in terms of Joules (1 J = 1 N * m). But it is often more useful to use the electron volt (symbol eV), which is the amount of work necessary (or the amount of energy acquired or lost)in moving an electron through a potential of one volt.

Now, V really means

$$V=V_b-V_a=E_B{r}_B-E_A{r}_A$$

However, from last week, we had

$$E=\frac{F_e}{q}=\frac{(\frac{\mathrm{kqQ}}{r^2})}{q}=\frac{\mathrm{kQ}}{r^2}$$

So we can write

$$V=k\frac{Q}{r_1}-k\frac{Q}{r_2}=k\left(\frac{Q}{r_1}-\frac{Q}{r_2}\right)$$

The work to move a charge q across a potential V difference in a field caused by a point charge Q will be

$$W=\mathrm{qV}=\mathrm{qk}\left(\frac{Q}{r_1}-\frac{Q}{r_2}\right)$$

Voltage or potential has meaning only as a difference, and we are free to choose the difference for any two points we like when we consider the electric potential around a point charge. If we choose r₂ to be at infinity, than the contribution to V of Q/r₂ goes to zero, and we can write:

$$V=k\frac{Q}{R}$$

Dipoles

This works only for single point charges. We have to make other considerations for dipoles, or situations where we have two equal charges Q of opposite signs separated by a short distance l (such as exists for the electron and positive nucleus of an atom, or the positive and negative sides of a polar molecule like the water molecule). At any point P from the dipole, the distance to one charge will be r and the distance to the second charge will be r + r', where r' may be as little as 0 if P lies on the line bisecting l, or as much as l if P lies on the extension of l into space. The distance r' is often given as l cos θ where θ is the angle at the further charge between its distance r + r' and l. (See diagram on p. 488 of your text). The total potential difference at P is the sum of the potentials caused by both charges, so

$$V=k\left(\frac{Q}{r}-\frac{Q}{\left(r+r\text{'}\right)}\right)=\mathrm{kQ}\left(\frac{1}{r}-\frac{1}{\left(r+r\text{'}\right)}\right)$$

We can rearrange the distance factors in parentheses if we multiply each factor by the denominator of the other (standard algebra trick):

$$\left(\frac{1}{r}-\frac{1}{\left(r+r\text{'}\right)}\right)=\frac{\left(r+r\text{'}\right)}{r\left(r+r\text{'}\right)}-\frac{r}{r\left(r+r\text{'}\right)}=\frac{r\text{'}}{r\left(r+r\text{'}\right)}$$

Since r' = l * cosθ, we can now write $$V=\mathrm{kQ}\left(\frac{l\cos \theta }{r(r+r\text{'})}\right)$$

Now we use the standard physicist's trick of placing P far from the dipole, so that r >> r' and we can effectively ignore r' in the denominator. Our equation for the dipole becomes

$$V=\mathrm{kQ}\left(\frac{l\cos \theta }{r(r+0)}\right)=\mathrm{kQ}\left(\frac{l\cos \theta }{r^2}\right)$$

Notice the interesting outcome of this: V with respect to a single point charge is proportional to 1/r, but V with respect to two close equal and opposite charges is proportional to 1/r².

Practice with the Concepts

Discussion Points

• If two points are at the same potential, does this mean that no net work is done in moving a test charge from one point to the other? Does this imply that no force must be exerted?
• An electron is accelerated from rest by a potential difference of V. How much greater would its final speed be if it is accelerated with four times as much voltage?

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