# Electric Fields and Forces

1. Apr 9, 2009

### km7

1. The problem statement, all variables and given/known data

Two negative charges of -1.0x10^-6 C each are placed in water and separated by a distance of 0.10m. A positive charge of 1.0x10^-8 C is placed exactly midway between the two negative charges. Determine:

a) the electric field (magnitude and direction) and electric potential at the position of the positive charge

b) the electric field (magnitude and direction) and electric potential at the position of either negative charge

c) the electric force (magnitude and direction) experienced by the positive charge

d) the electric force (magnitude and direction) experienced by either negative charge

2. Relevant equations

E = kq/Kr^2
V = kq1/Kr1 + kq2/Kr2
F = Eq ?

3. The attempt at a solution

let q1 be negative charge on the left and q2 be negative charge on the right
let q0 be positive charge that is in between q1 an q2

a)
E total = E1 + E2
= (- kq1/Kr1^2) + (+ kq2/Kr2^2) = 0 N/C
V = kq1/Kr1 + kq2/Kr2 = - 4.47 x 10^3 V

b)
E total = E0 + E2
= (+ kq0/Kr0^2) + (+ kq2/Kr2^2) = + 1.1626 x 10^4 N/C (direction ???)
V = kq0/Kr + kq2/Kr = - 1.09 x 10^3 V (first r = 0.05m, second r = 0.10 m ???)

c) and d)
F = Eq ?
F = kq1q1/Kr^2 ?

I tried solving them and got answers to a and b but not to c and d. I am confused when to include - or + sign in front of E, V, F, and q (do not understand what those signs mean exactly .. :[ ). Maybe I am not understanding the question correctly?

It would be great if anyone can explain them to me (basic concepts).

Thank you!

Last edited: Apr 9, 2009
2. Apr 9, 2009

### tiny-tim

Welcome to PF!

Hi km7! Welcome to PF!
Yes, F = qE is the correct equation …

the "Lorentz force" of an electric field E on a charge q is qE

in other words, the force is parallel to the field, and is multiplied by q (so if q is positive, then it's in the same direction, while if q is negative, then it's in the opposite direction)