equipotential lines and field lines; capacitance

Physics II

Class Notes, 4/04/99

In an electrostatic field, such as that of the two charges shown below, a test charge can be moved in a direction perpendicular to the electric field without any work being done by or on the charge.

Recall that work is equal to the product of the force in the direction of motion and the displacement.
Since motion perpendicular to the direction of the electric field will be perpendicular to the force, there will be no work associated with this motion.

The curve in the figure below is not a badly drawn circle. This curve is drawn so that at every point it is perpendicular to the electric field.

Between the charges, the electric field is effectively 'pulled' toward q2 and 'pushed' away from q1.
This causes the relative flattening of the curve perpendicular to the field in the region between the charges.

The curve drawn in this figure is called an equipotential curve.

Along an equipotential curve, a test charge can be moved with 0 work involved.

This implies that the voltage, which is the work per unit charge, is the same between any two points of the curve.
If we were to move from one point on an equipotential curve to another, without staying on the equipotential curve, the net work done would be zero.
It follows that along some parts of the path the field would do positive work on the charge and along other parts the field would do negative work.

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We have written a QBASIC program to plot field lines and equipotential lines. The program is instructive, and will be outlined here:

We begin by setting up the coordinates of the screen window and asking the user for the charges and the coordinates of the charges, then plotting two circles representing the charges:

WINDOW (-13, -10)-(13, 10)

PRINT "Screen dimensions are -13 to +13 horizontally and -10 to 10 vertically."

INPUT "charge 1: ", q1
INPUT "coordinates of charge 1: ", x1, y1
INPUT "charge 2: ", q2
INPUT "coordinates of charge 2: ", x2, y2

CIRCLE (x1, y1), .05
CIRCLE (x2, y2), .05

We proceed to plot the equipotentials and the fieldlines.

chois = 2

**chois tells the plotting routine whether to plot fieldlines or equipotentials (chois=2 plots equipotentials, chois = 1 plots field lines) **

FOR xx = -13 TO 13 ** xx is the x coordinate on the screen; y is just below 0 **

y = -.00001
x = xx
PSET (x, y) ** (x,y) is the starting point for the equipotential**
CALL fieldLine ** routine will plot an equipotential**

NEXT

**we set chois=1 so the plotting routine will know to plot fieldlines**
chois = 1
q = q1

** from a starting small angle to `pi get starting points for 24 field lines around q1, then plot the field lines below the y axis from each starting point**
FOR theta = 3.14 / 48 TO 3.15 STEP 3.14 / 24

** the point (x,y) will lie at distance 1 from the charge in the direction -theta from the charge **
x = x1 + COS(theta)
y = y1 - SIN(theta) ** - sin(theta) puts us below the y axis **
PSET (x, y)
CALL fieldLine ** this routine will plot the field line or equipotential from the starting point **
NEXT

** do likewise for q2 **
q = q2
FOR theta = 3.14 / 24 TO 3.15 STEP 3.14 / 24 * ABS(q1 / q2)
x = x2 + COS(theta)
y = y2 - SIN(theta)
PSET (x, y)
CALL fieldLine
NEXT

Video Clip #04

The plotting routine is as follows:

SUB fieldLine
SHARED x, y, x1, y1, x2, y2, q1, q2, chois, q ** these variables are shared with the main routine **

DO UNTIL xq = 999

** the following if-then-else routine cuts off the plotting of the fieldline; you don't need to understand it unless you want to learn to program**

IF choise = 1 THEN
    IF ABS(x) > 13 THEN xq = 999
    IF ABS(y) > 10 THEN xq = 999
ELSE
    IF ABS(x) > 13 THEN xq = 999
    IF ABS(y) > 10 THEN xq = 999
IF y > 0 THEN xq = 999
END IF

** We find the field strength and direction for the field at (x,y) due to q1 **
r1x = x1 - x
r1y = y1 - y ** x and y distances from (x,y) to charge **
r1 = SQR(r1x ^ 2 + r1y ^ 2) **Phythagorean Theorem for distance to charge **

sine1 = r1y / r1
cosine1 = r1x / r1 ** sine and cosine of angle from charge to field point (x,y) **
fieldStrength1 = q1 / r1 ^ 2 **Coulomb's Law **

** In the same way we find the field strength and direction for the field at (x,y) due to q1 **
r2x = x2 - x
r2y = y2 - y

r2 = SQR(r2x ^ 2 + r2y ^ 2)
sine2 = r2y / r2
cosine2 = r2x / r2
fieldStrength2 = q2 / r2 ^ 2

**We find the magnitude and direction of the total field at (x,y)**

fieldX = fieldStrength1 * cosine1 + fieldStrength2 * cosine2
fieldY = fieldStrength1 * sine1 + fieldStrength2 * sine2 ** x and y components of total field **
fieldStrength = SQR(fieldX ^ 2 + fieldY ^ 2) ** Pythagorean Theorem for vector magnitude**
sine = fieldY / fieldStrength
cosine = fieldX / fieldStrength ** sine and cosine of angle made by field at (x,y) **

SELECT CASE chois
    CASE 1:

        ** for field line, move (x,y) .1 unit in direction away from charges **
        x = x - .1 * cosine * q / ABS(q)
        y = y - .1 * sine * q / ABS(q)
    CASE ELSE:

        ** for equipotential, move (x,y) 1 unit in direction perpendicular to field line          (note negative reciprocal slope) **
        x = x + .1 * sine * q / ABS(q)
        y = y - .1 * cosine * q / ABS(q)
END SELECT

LINE -(x, y), 0 ** plot from last point to new point **

You can run the program, which is named FIELDS2, by clicking on its name.

Video Clip #05

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The figure below depicts the electric field lines emanating from two equal and unlike charges.

Think of the field as directed always from the positive charge on the left and toward the negative charge on the right.
The positive charge on the left will tend to 'push' the field away from it while the negative charge on the right will tend to 'pull' the field toward it.
The strength of the push or pull is inversely proportional to the square of the distance from the corresponding charge, and proportional to the magnitude of that charge.
We see that at points equidistant from the two charges, the push and the pull are equal, while at points closer to one charge or the other the influence of the closer charge dominates.

The 'flattened circles' are at every point perpendicular to the field lines (note that the aspect ratio in the figure below might result in a slight distortions of these angles).

These 'flattened circles' are the equipotential lines, since motion perpendicular to the field involves no work and hence no change in electrostatic potential.

The figure below depicts two equal and unlike charges.

We see that the 'push' of the positive charge on the left dominates the 'pull' of the smaller negative charge on the right except at points close to the smaller charge.

The figure below shows these two unequal charges as seen from a greater distance.

In this figure it is clear how the region in which the smaller charge dominates is localized in the vicinity of that charge.
At a great distance, the two charges act effectively like a single point charge equal to the sum of the two charges.
We can see that at a great distance the electric field is nearly uniform (i.e., the spacing of the field lines is nearly uniform), and that the direction of the field is nearly radial.
We see that the equipotential lines at the greater distances are very nearly circular.

In the figure below we see equal like charges.

We can think of the field lines as being equally pushed away from two equal positive charges.
Alternatively we can think of the field lines is being equally pulled toward two equal negative charges.
We see how the density of the field lines, and hence the strength of the electric field, is greater in the region between and below the two charges.
The field lines at the point halfway between the two charges repel one another in such a way that no field line can approach this point; this is consistent with the fact that the electric field at this point is 0.

At large distances the field due to these two charges will become more and more nearly radial, approaching that of a doubled charge at the midpoint between the two charges.

The equipotential lines will become more and more circular and field lines more and more uniformly spaced.

In the figure below we see unequal like charges.

Note the point between the charges at which the electric field is 0. The field lines are effectively repelled from this point.
Try to see how, at much greater distances, the field lines will become nearly radial and uniformly distributed.

The figure below depicts a capacitor consisting of two parallel plates, each with charge density +-`sigma.

Recall that each plate gives rise to a field 2 `pi k `sigma, with the fields directed so that they reinforce between the plates and cancel outside the plates.
The resulting field between plates is thus 4 `pi k `sigma.

The voltage across the plates is the work per unit charge required to move a test charge from one plate to the other.

Since the force on a test charge Q is F = Q * E = 4 `pi k `sigma Q, then the work done will be `dW = 4 `pi k `sigma Q d, where d is the distance between the plates.
The voltage is then the work per unit charge, which we obtain by dividing the work by the test charge, as seen in the last line of figure below.
We thus see that the voltage across the capacitor is 4 `pi k `sigma A.

We can find the charge to voltage ratio, or the capacitance, on a capacitor consisting of parallel plates of area A, with uniform separation d.

If the capacitor contains charges +- Q on its respective plates, then the charge density is `sigma = Q / A.
The voltage is therefore V = 4 `pi k `sigma d = 4 `pi k (Q / A) d.
It follows that the capacitance, which is the ratio Q / V, is

C = A / 4 `pi k d.

Video Clip #07

We thus see that the capacitance of a capacitor is directly proportional to the area of its plates and inversely proportional to the separation between the plates.

Thus, for example, to double the capacitance of a capacitor we could double the area of its plates, or we could halve the separation of the plates.

If we make a capacitor out of two pieces of aluminum foil, with a piece of paper sandwiched between them to keep the charges apart, we can estimate the capacitance of the system.

A standard piece of notebook paper has an area of approximately 600 cm² and a thickness of approximately .01 cm = .0001 m.
If this piece of paper is sandwiched between two pieces of aluminum foil of the same length and width as the paper, then we have a capacitor with area A = 600 cm² = .06 m² and thickness .0001 m.
For this capacitor A / d = 600 m, so that

C = (1 / 4 `pi k) * A / d

= 9 * 10^-12 C² / (N m²) * 600 m

= 5.4 * 10^-9 C² / J

= 5.4 * 10^-9 C / (J/C)

= 5.4 * 10^-9 C / V

= 5.4 * 10^-9 Farad.

(note that 1 / (4 `pi k) = 1 / (1.1 * 10^11 N m² / C²) = 9 * 10^-12 C² / (N m²), approx. )