# Homework Help: Question about electric fields

1. May 8, 2013

### Nerkiot

Hey guys, just got a question at school and the last part makes no sense. I replicated it almost exactly, except for the diagram. Just imagine the dotted line is a solid line indicating the electric field lines. Ignore the 4 dots around X, it was just to keep the field lines in line.

Question 4:
A section of a uniform electric field is represented by the field lines shown below:

.. ------------->-------------
X ------------->------------- Y
.. ------------->-------------
The section from X to Y is 8cm long and its electric potential difference is 50 V.

(a) Calculate the electric field strength and which side is positive.

E = V/d = 50/0.08 = 625 Vm^-1

X is the positive side as field lines come out of the positive and into the negative.

(b) How much energy would be needed to move a 140 micro-coulomb charge through this section of field from X to Y?

E = qV = 50 x (140 x 10^-6)
= 7 x 10^-3 J

(c) Deduce the nature of charge (positive or negative) from the information in the question and explain how you arrived at your answer.

Now first of all the question seemed ridiculous because it said it had a charge of 140 micro-coulombs, indicating a positive charge. However, after thinking about it, since the energy calculated in section (b) is equal to work done (V = W/q anyways), and if work needs to be done then the charge must be going against its normal path, this would indicate the charge was negative. In fact the answers state the charge IS negative as work needed to be done to move it from X to Y. This makes sense, since a negative charge would normally want to go to the X side as it is attracted to a positive side.

However, this work value was determined by substituting a positive value of charge into E = qV. I am confused by this inconsistency - a positive charge would not need work to move from X to Y, but when you substitute it into E=qV you will naturally get a positive value, meaning work needed to be done. The only explanation I can come up with this is that the value of "V" I needed to substitute in needed to be negative, but I see no reason as to why.

Thanks for any help!

2. May 8, 2013

### Andrew Mason

This is just a matter of sticking to the convention.

The convention is that the electric field direction is the direction of the electric force on a unit of positive charge. So if positive work has to be done by a charge in moving from X to Y (ie. the work is done by an external force acting on the charge against the electric field), the electric force on the charge is from Y to X (ie. opposite to the direction of the electric field). So the charge is negative.

The expression W = qV = qEd gives the work done ON the charge by the electric field. W=-qV = -qEd gives the work done BY the charge against the electric field. In this case, since q is negative, W is positive (ie. positive work is done BY the charge against the electric field).

AM

Last edited: May 8, 2013