# Negative charge - To take or not to take an abs val

## Homework Statement    See above

## The Attempt at a Solution

In the solutions guide for problem 71, if a charge is negative (look at the -1.5 C charge), it will be negative in the equation for potential energy (last step where multiplication brings 1.5J. On the other hand, in question 26, the solution guide takes the absolute value. Is there a general guide to when I should take the absolute value of a negative charge or if I should plug in the negative number into the formula. Thank you

gneill
Mentor
Since the field lines of electric charge radiate radially from a point charge, the directions of the field lines depend upon your location with respect to the charge. It can be pretty complicated trying to accommodate the effect of a change of position on the field direction in the mathematics of a single equation. Usually the most expedient thing to do is sketch the setup, pencil in the field directions, pick a point and then write the equations using the absolute values of the charges and incorporating the "direction" information in the signs between terms.

I can't think of anything that could be considered a general guiding principle here. If you want to wrestle with the mathematics of each problem to accommodate all the nuances of the geometry you can do so. It tends to over complicate things though, and is akin to debugging a program to get it sorted out. Using magnitudes and writing expressions for particular locations is easier to do on the fly.

An easier way to understand would be that the "-" sign is an indicator of direction. Just like the "-" sign in Faraday's law of electragnetic induction is an indicator of "direction" of current.

Question 26 is clear - the fields due to the two charges cancel. Hence as we have an idea of the direction in our head already ( field by a negative charge is radially inward towards it and vice versa for positive charge), we can just proceed with absolute values. But all this should be done knowing that we have the direction of field in the back of our heads.

Q 71, on the other hand, requires a value of PE which depends on the charge. If we put the modulus of the value in, then we must physically draw field lines and verify that Potential actually drops or increases. The substitution of the charge as it is bypasses this process and directly gives us the answer.