posted by wendlen on .
Consider three charges q1 = 6.0 µC, q2 = 1.8 µC, and q3 = -2.8 µC, arranged as shown below.
Q1 is positive and is 3 cm away from Q2. Q2 is also positive and is 2 cm away from Q3 which is negative.
(a) What is the electric field at a point 1.0 cm to the left of the middle charge?
(b) What is the magnitude of the force on a 1.1 µC charge placed at this point?
Electric field at a distance r from a charge q is given by Coulomb's law:
E = k q / r^2
To get the E-field, you have to add the contribution of 3 charges:
E1 = k q1 / r1^2
E2 = k q2 / r2^2
E3 = k q3 / r3^2
You can look at the geometry and so simple addition and subtraction to get the three r's. They give you the three q's. Don't worry about sign just yet. Let all the numbers above be positive.
Then you add them up, but you have to be a little careful nowabout direction. Let's call right positive. q1 is to the left of the point and positive. Positive charges push the E-field away, so it's to the right, and positive. q2 is to the right of the point and positive, so it pushes left (negative). q3 is to the right and negative, so it pulls right (positive).
So Etotal = E1 - E2 + E3 (positive answer means to the right)
And of cource, the force of that E-field on a test charge q, is just:
F = qE