Can someone please help me with this physics question?

M

Megan

New member
Two small electrified objects A and B are separated by a 0.03 meter, and repel each other with a force of 4.0 X 10 to the negative five newton. If we move body A an additional 0.03 meter away, what is the electric force now?
 
Alright you get ONE answer from me today :D.

ok so the force acting on these objects is the Coulomb force ... this is given by the formula

F = K Qq/r^2

how can we use this to get the difference between the two different forces ... simple we have two forces and two different distances so let's call them F1 and F2 ... which correspond to R1andR2.

F1 = K Qq/R1^2
F2 = K Qq/R2^2

ok so we can divide these two formulas ... this gets rid of the KQq terms (pretty neat eh?)

F1/F2 = 1/R1^2 /(1/R2^2)
F1/F2 = R2^2/R1^2

now we just solve for F2

F1 * (R1^2/R2^2) = F2
F2 = 4.0E-5 * (0.03^2/0.06^2)
F2 = 1.0E-5 N

you're welcome!
 
Fnet = k*q1*q2/ r^2 = 4*10^-5 newtons


so if we increase r by 0.03 meters then all we are really doing is doubling the distance from r to 2r:
Fnet will decrease to k*q1*q2/ (2r)^2

or by a 1/4 since 1/(2*r)^2 = (1/4)*(1/r^2)

so the electrical force will be one quarter of what it was originally:
=(1/4)*4*10^-5 newtons
= 1*10^-5 newtons

This problem is designed to show that when you double the distance the force falls to a quarter of what it was. This is because the electrostatic force uses an inverse square of the distance between to charged points. This is the same way gravity works as well.

hope that helps!
 
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