## SAT Physics Subject Test

**Chapter 9 ****Electric Potential and Capacitance**

**ELECTRIC POTENTIAL**

Consider a point charge *q*_{1} at a distance *r*_{1} from a fixed charge *q*_{2} moving by some means to a position *r*_{2}

How much work did the electric force perform during this displacement? The answer is given by the equation *W*_{E} = −*kq*_{1}*q*_{2} .

Therefore, since Δ*U** _{E}* = –

*W*

*we get*

_{E}*U*

_{2}–

*U*

_{1}=

*kq*

_{1}

*q*

_{2}.

Let’s choose our *U*=*U* reference at infinity. Then this equation becomes *U** _{E}* = .

Notice that if *q*_{1} and *q*_{2} have the same sign, *U*_{E} is positive. This means it took positive work by an external force to bring these charges together (since they repel each other). If *q*_{1} and *q*_{2} have opposite signs, *U*_{E} is negative. This means the electric force does positive work to bring these changes together (since they attract).

Let’s look at another example.

Let *W*_{E} be the work done by the electric field on a charge *q* as it undergoes a displacement. If another charge that’s twice as strong, say 2*q*, were to undergo the same displacement, the electric force would be twice as great on this second charge, and the work done by the electric field would be twice as much, 2*W*_{E}. Since the work would be twice as much in the second case, the change in electrical potential energy would be twice as great as well, but the ratio of the change in potential energy to the charge would be the same: . This ratio says something about the*field* and the *displacement*, but not the charge that made the move. The change in **electric potential**, *V*, is defined as this ratio.

Electric potential is electrical potential energy *per unit charge*; the units of electric potential are joules per coulomb. One joule per coulomb is called one **volt** (abbreviated V); so 1 = 1 V.

Consider the electric field that’s created by a point source charge *Q*. The electric potential at a distance *r* from *Q* is

*V* =

Notice that the potential depends on the source charge making the field and the distance from it.

3. Let *Q* = 2 10^{–9} C. What is the potential at a point P that is 2 cm from *Q* ?

Here’s How to Crack It

Relative to *V* = 0 at infinity, we have

Notice that, like potential energy, potential is a *scalar*. In the preceding example, we didn’t have to specify the direction of the vector from the position of *Q* to the point P because it didn’t matter. At *any* point on a sphere that’s 2 cm from *Q*, the potential will be 900 V. These spheres around*Q* are called **equipotential surfaces**, and they’re surfaces of constant potential. Their cross sections in any plane are circles and are, therefore, perpendicular to the electric field lines. The equipotentials are always perpendicular to the electric field lines.

4. How much work is done by the electric field as a charge moves along an equipotential surface?

Here’s How to Crack It

If the charge always remains on a single equipotential, then the potential, V, never changes. Therefore, Δ*V* = 0, so Δ *U*_{E} = 0. Since *W*_{E} = –Δ *U*_{E }, the work done by the electric field is zero.

5. Two large, flat plates—one carrying a charge of +*Q*, the other –*Q*—are separated by a distance *d*. The electric field between the plates, **E**, is uniform. Determine the potential difference between the plates.

**Potential vs.Potential Energy**

Remember that everything

naturally tends toward

lower potential energy.

Positive charges tend

toward lower potential

and negative charges tend

toward higher potential.

Here’s How to Crack It

Imagine a positive charge *q* moving from the positive plate to the negative plate.

Since the work done by the electric field is

*W*_{E,+→−} = F_{E}*d* = *q*E*d*

the potential difference between the plates is

This tells us that the potential of the positive plate is greater than the potential of the negative plate, by the amount E*d*.