Chapter 17
The Giancoli onLine Tutor

Electric Potential and Electric Energy; Capacitance

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I. The Constant Electric Field

Finding the Speed of a Charged Particle in a Capacitor.

The figure at the left shows the capacitor from the previous chapter. The red dot represents a positive charge, q, which is free to move. The only force acting on it (ignoring gravity) is the electric force (F=qE). Since the Electric Field is constant (uniform) in both magnitude and direction, so is the force. Any problem involving the motion of a positively charged particle in the interior of a capacitor is exactly like the motion of an object near the surface of the earth. Near the Earth gravity is taken to be given by F_g=mg , for a charge inside a capacitor the force is given by F=qE. "q" plays the role of "m" and "E" plays the role of "g". The gravitational potential is given by PE=mgy where y is normally taken to be the height above the ground. The choice of where the zero is taken is arbitrary and chosen for convience. Similarly, we can take the zero of the electric potential energy to be the negative plate of the capacitor, and, then, by analogy we discover that the potential energy of the positive plate is given by PE=qEd. Problems will be worked the same way, for Example,

A mass, m, is released from a height h, how fast is it going when it hits the ground?

Using Conservation of Energy:

Energy Before= Energy After

mgh=mv²/2

v=SQRT(2mgh/m)

A positive charge, q, is released from the positive plate of a capacitor with separation d, how fast is it going when it hits the negative plate?

Using Conservation of Energy:

Energy Before= Energy After

qEd=mv²/2

v=SQRT(2qEd/m)

Interactive Example 1
Suppose a positive charge with a charge to mass ratio (q/m) of 2 C/kg is released at the positive plate, if E=10000 N/C and d =1 cm, what is its speed when it reaches the negative plate? Enter your answer below and click outside the box to get your result.

What would happen if there was a negative mass? Well, the potential energy at height h, mgh, would now be lower than the potential energy at the ground, because it is negative. Objects with a negative mass would 'fall' up! While objects of negative mass do not exist in our normal matter world (presumably), objects of negative charge do exist and they do indeed 'fall' up, as shown in the figure on the right where the blue dot represents a negative charge which will be accelerated towards the positive plate.

So how would we find the speed with which the negative charge strikes the positive plate? Obviously the above solution won't work because it would put a negative number inside a square root. The solution is given below:

A negative charge, q, is released from the negative plate of a capacitor with separation d, how fast is it going when it hits the positive plate?

Using Conservation of Energy:

Energy Before= Energy After

0=mv²/2+qEd

v=SQRT(-2qEd/m)

(remember q, itself, is negative, so the quantity in the square root is positive.)

The Electric Potential

While we, in our read world, can always say that the top of a tree always has a potential energy greater than its base, the same can not be said of the positive plate of a capacitor relative to the negative plate. For this and other reasons we define the Electric Potential or, simply, the Potential, or, informally, the Voltage by:

V=PE/q

The positive plate of the capacitor will always have a higher potential than the negative plate. The units of Potential are joules/coulomb and is known as a volt, abbreviated as V.

The units tell us why a 6 volt Gulf Cart battery is large and heavy and a 9 volt transistor battery can be small and light even though it has a higher voltage. The Gulf Cart battery might only transfer 6 J of energy for every coulomb of charge it transfers, but it can transfer tens of thousands of coulombs of charge while the 9 volt battery could only supply a very small fraction of this charge.

Work and the Electric Field

In the above discussion about the capacitor we assumed that the Potential Energy and Potential of the negative plate was zero, while this is convenient it is not necessary to do so. It can be any number, only differences in Potential Energy (and Potential) matter. So to be more general we take the potentials of the two plates as V_b and V_a, respectively. The Potential difference is, then,

V_ba= V_b-V_a
and the Potential Energy difference is

DPE=qV_ba.
If q is negative DPE will be negative and conversely.

Since the electric force is a conservative force, mechanical energy is coserved, which means that

DKE= -DPE

Since, ignoring the very small effect of gravity, the only force acting on the charge in the capacitor is the Electric Force, the work done by the Electric Force is the Total Work and must equal the change in Kinetic Energy. So

W_b->a= -DPE= -(PE_a-PE_b)= (PE_b-PE_a) =q(V_b-V_a)=qV_ba.
Or in brief,

W_ba= qV_ba.

In summary:

	b is positive plate	b is negative plate
q is positive	W_ba is positive (A positive charge is accelerated from positive to negative.)	W_ba is negative (A positive charge is 'pushed' from negative to positive)
q is negative	W_ba is negative (A negative charge is 'pushed' from positive to negative.)	W_ba is positive (A negative charge is accelerated from negative to positive.)

Of course Work can be calculated in other ways. Since The electric field is constant inside the capacitor, so will the force be constant. Thus
W=qEdcos(q),
where q is 0 if the charge is moving from "+" to "-" and is 180^o if the motion is from "-" to "+". Comparing the two formulae for work and we find that,
V_ba= Ed or E= V_ba/d
This is a very useful relationship for calculating the Electric field since the Voltage on the capacitor and the separation between the plates is usually known. For example, a capacitor with a plate separation of 1 cm and an applied voltage of 100 volts would have an internal electric field of
E= 100 V/.01 m= 10000 V/m.
Note the alternate units, V/m, for the electric field.

Interactive Example 2
a) In example 1, what is the potential difference between the plates of the capacitors? Enter your result below.

b) If the mass of the charge in example 1 is 10 grams what is the work done by the electric field. Enter your result below.

II. The Electric Potential and the Point Charge

Another case where we use the electric potential is with point charges. The formula for the potential at a point 'x' a distance R from a charge Q is

V=kQ/R

Interactive Example 3
In the figure on the left, the blue charge is negative and the red charge is positive; each charge has magnitude of (1/9)x10^-8 C so that kQ equals 10 Vm for each charge.

1) What is the Potential at the point 'x' from the blue charge?

2) What is the Potential at the point 'x' from the red charge?

3) What is the Potential at the point 'x' from the red charge?

Interactive Example 4

(To use this resource the Quicktime3 or Quicktime4 plugin get it here: )
In the above 'movie' you can drag the blue indicator bar horizontally back and forth. The number given in the upper left hand corner is the Voltage (Volts) at the location. The green charge is preset to 1 mC but the red and blue charge are unknown and it is your 'job' to determine them. This can be done mathematically or you can do it experimentally if you are clever enough. To set the charge values for the blue and green charges, click on them. Do it now! All charges are in microcoulombs and are in the range -10 to 10 with no decimal places ( i.e., -10, -9, -8, ... 10).

When you think you know the answers

enter them below

and then

click the GradeIt button.

The electric potential outside of a charge (given in microcoulombs) is given by

V= 9000*Q/Dx

where
Dx
is the distance from the center of a charge to the position x where the voltage is measured. The distance scale along the bottom is in meters and the center of the charges are at 50, 200, and 350, respectively. Note: You cannot measure the voltage inside the charge!

Thus the voltage from the 3 charges is given by

V= 9000*Q₁/Dx₁ + 9000*Q₂/Dx₂ + 9000*Q₃/Dx₃
where all charges are in microcoulombs. (That is why the constant is 9000 instead of 9x10⁹)
To find Q₂ and Q₃, move the pointer to a known position and record the reading, for example, 1400 V. Then set up the above equation, by pluging in the distances (remember you know Q₁) and getting it in the form

k+lQ₂+mQ₃=1400,

where k, l and m are numbers. Next change the position of the pointer and repeat the above. You now have two equations with two unknowns. You can solve them for Q₂ and Q₃. If you are successful with finding them you will be given a code for each one, send the codes to me.
Enter your answer in mC for The Red Charge here:
Enter your answer in mC for The Blue Charge here:
and then
Your results will be here

The Capacitor and Energy

The amount of charge, Q on a capacitor is proportional to the voltage, V, across it. The proportionality constant, C, is called the Capacitance. Its unit is the farad, abbreviated as F. For a simple Capacitor

C=e_oA/d,
where A is the area of either plate and d is the separation between the plates; e_o is the permittivity of free space. If there is material between the plates (which normally there is), e_o is replaced by e= Ke_o, where K is the Dielectric Constant. K will normally be given!

If the plates are 10 cm by 10 cm the area is .01 m². If the separation is 1mm the capacitance is given by

8.85x10^-12 .01 =8.85x10^-11 F

.001

If 12 volts is applied to this capacitor the charge on it would be Q= 8.85x10^-11x12= 1.1x10^-9 C. The energy, U, stored in the capacitor is given by

U= QV/2= CV²/2= Q²/2C,

which means that the above capacitor has stored energy of 6.4x10^-9 joules. (Check This By Using Each of the Three Formulae)