Page 286 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 23] ELECTRICITY 271

SOLVED PROBLEM 23.12
How strong an electric field is required to exert a force on a proton equal to its weight at sea level?
The electric force on the proton is F = eE, and its weight is mg. Hence eE = mg, and
2
mg (1.67 × 10 −27 kg)(9.8 m/s ) −7
E = = = 1.02 × 10 V/m
e 1.6 × 10 −19 C

POTENTIAL DIFFERENCE
The potential difference V between two points in an electric field is the amount of work needed to take a charge
of 1 C from one of the points to the other. Thus
W
V =
q
work
Potential difference =
charge
The unit of potential difference is the volt (V):
joule
1 volt = 1
coulomb
The potential difference between two points in a uniform electric field E is equal to the product of E and
the distance s between the points in a direction parallel to E:
V = Es

Since an electric field is usually produced by applying a potential difference between two metal plates s apart,
this equation is most useful in the form
V
E =
s
potential difference
Electric field =
distance
A battery uses chemical reactions to produce a potential difference between its terminals; a generator uses
electromagnetic induction (Chapter 28) for this purpose.

SOLVED PROBLEM 23.13
6
The potential difference between a certain thundercloud and the ground is 2 × 10 V. Find the energy
dissipated when a charge of5Cis transferred from the cloud to the ground in a lightning stroke.

6
7
W = qV = (5C)(2 × 10 V) = 1 × 10 J
This is not a lot of energy; burning a cup of gasoline liberates more energy than this.

SOLVED PROBLEM 23.14

A potential difference of 20 V is applied across two parallel metal plates, and an electric field of 500 V/m
is produced. How far apart are the plates?
Since E = V/s, here
V 20 V
s = = = 0.04 m = 4cm
E 500 V/m