How much work must be done to move charge Q2 = 2.0 x 10-6 C .....

Question

How much work must be done to move charge Q2 = 2.0 x 10-6 C from A to B? The other charged object has a charge Q1 = 8.0 x 10-6 C.

 

                              5.0m                                       2.0m

+<------------------------------------------>+<------------------------->+

                                                           B                                     A

Q1=8*10^-6 C                                                                             Q2=2*10^-6 C

 

 

Answer: 8.2*10^-3 Joules

 

 

 

Answer 1

The amount done to move Q₂ from A to B is given by:

 

W = Q₂V_AB

 

where V_AB is the change in electric potential (voltage) between points A and B.  I assume that the Q₁ is fixed (as is often the case in these problems), since, otherwise, we do not have enough information (such as the speed with which the particle is moved) to solve the problem.

 

If we assume Q₁ is stationary (making Q₂ a "test charge," which is only acted upon and does not act on other particles), then V_AB is the difference in electric potential due to Q₁ at point B and at point A.

 

The expression for the electric potential a distance r from a point charge q is:

 

V(r) = (kq/r)

 

Note that, since electric potential is a scalar, you can keep the sign on the charge and get the appropriate sign for voltage from this expression (unlike for electric fields and forces, which are vectors, because the sign of a vector means something different than the sign of a charge).

 

So, in this case, V_AB = V_B - V_A = (kQ₁/r_B) - (kQ₁/r_A) = kQ₁((1/r_B) - (1/r_A))

 

Note that because Q₁ is a positive charge, and r_B < r_A, V_AB should be a positive number.  We expect the change in potential to be positive as we move closer to a positive charge since, by definition, electric potential is higher near positive charges.

 

Then the work W_AB = Q₂V_AB

 

If I use this method, I obtain the answer you list.  Note that the work we do is positive, corresponding to pushing the positive charge Q₂ a way it does not want to go (toward the other positive charge).  Doing positive work against a conservative force like the electric force will store positive electric potential energy (equal to the amount of work we do).  It is like we pushed that positive charge up a potential "hill," and the charge gained electric potential energy as a result.

 

If you have any more questions about this, just let me know.  I am also available for online tutoring, if you would ever like.