Electric Field Strength as a Vector along a line

[mickwess]

When Calculating Electric Field Strength using Q/kr^3, if the Q is negative is this negative disregarded? If not this would make E around a negative charge, negative. But if a positive test charge (q) is placed between two charges (Q1(negative) and Q2(positive)) then for Q1 (the negative charge) F will be negative becasue of the equation F=qE, and for Q2, F will be positive. The resultant of these forces is to be found by F1+F2, but this would mean that the resultant force F, would be smaller? And with one F being negative and the other not this suggests they act in different directions, which they dont!

 

[L-x]

E is a vector. Its direction is the direction of the force it will produce on a positively charged particle, in the case of the field due to a point charge no, you do not ignore the sign of the charge, but the expression you have is slightly wrong. It also involves the vector r, measured from the point charge.

Try drawing the problem and you should see where you are going wrong, you are measuring r from two different places.

 

[mickwess]

But one F will be positive and the other negative (they act the same way)

-0------------+q----------+0
Q1 Q2
<<<<<Both forces act this way<<<<

 

[L-x]

You are defining positive F to be the direction away from the particle in each case. This makes no sense if you are then trying to add the forces together in the way that you have done, because each force is defined relative to a different origin.

In this 1d case it makes more sense to say positive means pointing right, negative means pointing left, and thinking about where each F will be positive and where it will be negative.