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Solution to Problem 20.54 in CPHY 122: Electric Charge Conservation and Calculation, Assignments of Physics

The solution to problem 20.54 in the cphy 122 course taught by h. L. Neal. The problem deals with the conservation of electric charge and the calculation of charges on two identical spherical conductors when they are brought together. The fundamental principle of charge conservation, derives the equation for the charge on each sphere, and provides the solutions for the charges. Useful for students studying electrostatics and electric charge.

Typology: Assignments

Pre 2010

Uploaded on 08/04/2009

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CPHY 122
Solution to Problem 20.54
Instructor: H. L. Neal
January 30, 2009
The conservation of electric charge is essential to solving this problem. One statement
of this fundamental principle is that the total charge of a closed system is always constant.
Here the system is represented by the two identical spherical conductors. Initially we have
the total charge as q1+q2. When the two spheres are brought together, the total charge is
rearranged, so that each sphere has the same charge q. This gives the โ€ฆnal total charge as
2q. Conservation of charge requires the q1+q2= 2q. Since the magnitude of the repulsive
force is
F=kq2
r2;
the charge qis
q=rpF=k:
For the attractive force of magnitude Fwe must have q1q2<0, so that
q1q2=๎˜€q2:
This last equation may be transformed into the quadratic equation for q1(or q2) by using
the conservation of charge equation to write q2in terms of q1:
q1(2q๎˜€q1) = ๎˜€q2:
Moving everything to the right-hand side of the equation gives
q2
1๎˜€2qq1๎˜€q2= 0:
The solutions for q1are
q1=1
2๎˜2q๎˜†p4q2+ 4q2๎˜‘
=q๎˜1๎˜†p2๎˜‘:
It should be easy for you to show that
q1=q๎˜1 + p2๎˜‘;
q2=q๎˜1๎˜€p2๎˜‘:
1

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CPHY 122

Solution to Problem 20.

Instructor: H. L. Neal

January 30, 2009

The conservation of electric charge is essential to solving this problem. One statement

of this fundamental principle is that the total charge of a closed system is always constant.

Here the system is represented by the two identical spherical conductors. Initially we have

the total charge as q 1 + q 2. When the two spheres are brought together, the total charge is

rearranged, so that each sphere has the same charge q. This gives the ร–nal total charge as

2 q. Conservation of charge requires the q 1 + q 2 = 2q. Since the magnitude of the repulsive

force is

F = k

q^2

r^2

the charge q is

q = r

p F=k:

For the attractive force of magnitude F we must have q 1 q 2 < 0 , so that

q 1 q 2 = q 2 :

This last equation may be transformed into the quadratic equation for q 1 (or q 2 ) by using

the conservation of charge equation to write q 2 in terms of q 1 :

q 1 (2q q 1 ) = q 2 :

Moving everything to the right-hand side of the equation gives

q 2 1 ^2 qq^1 ^ q

2 = 0:

The solutions for q 1 are

q 1 =

2 q 

p 4 q^2 + 4q^2

= q

p 2

It should be easy for you to show that

q 1 = q

p 2

q 2 = q

p 2