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Equipotential Surface and Capacitors - General Physics II, Notes | PHYS 112, Exams of Physics

Material Type: Exam; Professor: Cattell; Class: General Physics II; Subject: Physics; University: Community College of Philadelphia; Term: Unknown 1989;

Typology: Exams

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p. 27
Equipotential Surfaces and Capacitors
Definition An equipotential surface is a locus of points that are all at the same electric potential.
Thus an equipotential surface is a surface on which the electric potential is the same everywhere. The
equipotential surfaces surrounding an isolated point charge are concentric spheres. See the figure below.
Taking the electric potential to be zero at infinity, the electric potential at
point D in the figure is
kq
r
. The electric potential at point B is lower than
that at point D because point B is farther away from the charge q.
If a small positive test charge q0 moves along the outer equipotential
surface along the path from A to B the work WAB done by the electric field
is given by Equation 19.4:
VB – VA = –WAB/q0
But VB = VA so that WAB = 0 and the electric field does no work on the test
charge. This is possible only if the electric force acts in a direction that is
perpendicular to the displacement of the test charge. (Remember the
formula for work from Physics 111: W = Fscos(), where is the angle
between the force F and the displacement s. If = 90 then W = 0 J.)
We conclude that the electric field lines intersect an equipotential surface at right angles. Note that the electric
field lines are directed from high electric potential to low electric potential.
Question: In the figure at the right, which of the three equipotential surfaces shown
is at the highest electric potential? Which is at the lowest?
There is a quantitative relationship between the electric field and equipotential
surfaces. Consider a region of space where there is a uniform electric field directed
from left to right. (Such a region exists, for example, near the center of a parallel-
plate capacitor. See the figure below.) In this region the equipotential surfaces are
parallel vertical planes.
If a small positive test charge q0 moves from
point A to point B the work done on the charge by
the electric field is given by
WAB = Fs
Where F is the magnitude of the electric force on q0:
F = q0E
E
E
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Equipotential Surfaces and Capacitors

Definition An equipotential surface is a locus of points that are all at the same electric potential.

Thus an equipotential surface is a surface on which the electric potential is the same everywhere. The

equipotential surfaces surrounding an isolated point charge are concentric spheres. See the figure below.

Taking the electric potential to be zero at infinity, the electric potential at

point D in the figure is

kq

r

. The electric potential at point B is lower than

that at point D because point B is farther away from the charge q.

If a small positive test charge q 0

moves along the outer equipotential

surface along the path from A to B the work W AB

done by the electric field

is given by Equation 19.4:

V

B

– V

A

= –W

AB

/q

0

But V B

= V

A

so that W AB

= 0 and the electric field does no work on the test

charge. This is possible only if the electric force acts in a direction that is

perpendicular to the displacement of the test charge. (Remember the

formula for work from Physics 111: W = Fs cos(), where  is the angle

between the force F and the displacement s. If  = 90 then W = 0 J.)

We conclude that the electric field lines intersect an equipotential surface at right angles. Note that the electric

field lines are directed from high electric potential to low electric potential.

Question: In the figure at the right, which of the three equipotential surfaces shown

is at the highest electric potential? Which is at the lowest?

There is a quantitative relationship between the electric field and equipotential

surfaces. Consider a region of space where there is a uniform electric field directed

from left to right. (Such a region exists, for example, near the center of a parallel-

plate capacitor. See the figure below.) In this region the equipotential surfaces are

parallel vertical planes.

If a small positive test charge q

0

moves from

point A to point B the work done on the charge by

the electric field is given by

W

AB

= Fs

Where F is the magnitude of the electric force on q 0

F = q 0

E

E

E

Thus

W

AB

= q 0

Es

But from Equation 19.4 V B

– V

A

= – W

AB

/q 0

so that

V

B

– V

A

=Es

or

E =  V / s

where  V = V B

– V

A

. The quantity  V / s is referred to as the potential gradient and has units of volts per meter.

Remarks. The potential gradient is a vector field that points in the direction of the most rapid rate of increase of

the potential. Because of the negative sign in the above formula we see that the electric field points in the

direction of the most rapid rate of decrease of the potential. (Proofs of these statements are beyond the scope of

this course. However, interested students may refer to a suitable calculus text such as Anton’s Calculus .)

Remark. Because of the above formula we see that alternate units for the electric field are volts per meter : V/m.

Exercise: Show that 1 V/m = 1 N/C.

Example

The plates of a parallel plate air capacitor are connected to the terminals of a 5.0 volt battery. The plates are

separated by a distance of 3.0 mm. Neglect fringing effects.

a. Find the magnitude of the electric field between the plates.

b. Find the surface charge density that must exist on each plate.

a.

3

3 V

m 3

; 5.0 V; 3.0 10 m

5.0 V

3.0 10 m

V

E V s

s

E E

b.

2

2

2 2

0

12 3 C V

0 0 m

N m

12 3 8 C C N C

C N m m

For a parallel plate air capacitor , where is the surface charge density on either plate. Thus

E

E E

   

  

Since E 0

E the dielectric constant  is a number greater than unity. See Table 19-1 in your text for dielectric

constants of some common substances.

TABLE 19.1 Dielectric Constants of Some Common Substances

a

Substance Dielectric Constant, k

Vacuum 1

Air 1.

Teflon 2.

Benzene 2.

Paper (royal gray) 3.

Ruby mica 5.

Neoprene rubber 6.

Methyl alcohol 33.

Water 80.

The Capacitance of a Parallel Plate Capacitor

Recall from the previous example that the magnitude of the electric field between the plates of a parallel plate

capacitor can be calculated from

or where and is the plate separation.

V V

E E V V d s

s d

Now E = E 0

/ and E 0

0,

where  is the surface charge density on either plate:  = q / A where A is the

surface area of each plate. Thus

0

0 0 0

0

0

or

E V V V V q V

E

d d d d A d

A

q V

d

A comparison of the last equation with q = CV shows that for a parallel plate capacitor

0

A

C

d

Note that from last formula above we see that  0

can be expressed in the units F/m. (Why?)

Example

Suppose an air capacitor consisted of two identical square copper plates placed side by side with a plate

separation of 1.0 cm. Find the cross-sectional area of each plate if the capacitance of the capacitor is to be 1.0 F.

 

0 0

0

2

9 2

12 F

m

; here 1 so that and.

1.0 F 1.0 10 m

; 1.1 10 m

A A Cd

C C A

d d

A A

Each of these plates would have a side over 33 km long! This example illustrates that, typically , common

capacitors that you hold in your hand have capacitances that are only small fractions of a farad.

Effect of a Dielectric on Capacitance

From the previous example we saw that if just air is between the plates of a parallel plate capacitor the

capacitance is

0 0

A

C

d

. When a dielectric with dielectric constant  is placed between the plates the

capacitance becomes

0

A

C

d

. We can conclude that, for the parallel plate capacitor,

0

C  C

It can, in fact, be shown that this formula is valid for all capacitors, regardless of their shape. So the insertion of

a dielectric between the plates of a capacitor has the effect of increasing the capacitance of the capacitor.

Remark. The use of a dielectric to increase capacitance has a drawback. When a dielectric is subjected to an

electric field that is strong enough the dielectric breaks down and becomes a conductor. The minimum electric

field strength to which a dielectric can be subjected without breakdown is called the dielectric strength of the

dielectric. The dielectric strength of air, for example, is 3  10

6

V/m; for paper it is 16  10

6

V/m. Placing a

dielectric between the plates of a capacitor, while increasing the capacitance, also limits the potential difference

that can be placed across the plates of the capacitor. If too high a potential difference is placed across the

capacitor’s plates the dielectric strength of the dielectric between the plates can be exceeded causing the

capacitor to break down and become useless.

Synthetic dielectrics have been developed that have enormous dielectric constants. A demonstration will be

given in class of capacitor with a cross-sectional area about the size of a quarter that has a capacitance of 1.0 F.

Example

a. Estimate the capacitance of a capacitor made from two quarters with a piece of notebook paper placed

between them.

b. What would the dielectric constant of the material placed between the quarters have to be to produce a

capacitance of 1.00 F? (Assume the same separation of the quarters.)