Coulomb's Law: Find q/Q so force is 1/3 maximum possible

S

soccerball3211

Junior Member
Joined
May 16, 2005
Messages
84
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby, sphere. Both spheres can be treated as particles. For what value of q/Q will the electrostatic force between the two parts have 1/3 of the maximum possible value?

This is what I did:

. . .F = (k*Qq - q^2) / r^2

I then differentiated the force with respect to charge and set the result equal to zero to find that Q = 2q or q = Q/2

I then plugged q into the original equation to find:

. . .Fmax = kQ^2 / (4r^2)

I can't figure out what to do now! Can anyone help?

Thank you!
 
q = Q/2 and Q = 2q is correct.

Fmax = kQ<sup>2</sup>/(4r<sup>2</sup>)

using 2q for Q ...

Fmax = k(4q<sup>2</sup>)/(4r<sup>2</sup>) = kq<sup>2</sup>/r<sup>2</sup>

so ... 1/3 of Fmax = kq<sup>2</sup>/(3r<sup>2</sup>)

kq<sup>2</sup>/(3r<sup>2</sup>) = k(Q - q)q/r<sup>2</sup>

q<sup>2</sup>/3 = (Q - q)q

since q is not 0 ...

q/3 = Q - q

4q/3 = Q

q/Q = 3/4
 
don't know what to tell you ... we both got the same initial solution, that should tell you something.
 
checked both 1/4 and 3/4 using actual values ... neither works.

I tried again ... took a different route that involved the quadratic formula and obtained this solution.

\(\displaystyle \L \frac{q}{Q} = \frac{1 - \sqrt{\frac{2}{3}}}{2}\)

This one works ... third time might be the charm.
 
You must log in or register to reply here.
Top