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## Homework Statement

Twelve equal particles of charge ##+q## are equally spaced over a circumference (like the hours in a watch) of radius R. At the center of the circumference is a particle with charge ##+Q##.

a) Describe the net force acting over ##+Q##.

b) If the charge located at "3'o'clock" has been taken off, what's the net force acting over ##+Q##?

c) If both charges located at "3'o'clok" and "9'o'clock" has been taken off, describe the force acting over +Q this time.

## Homework Equations

##\vec F= \frac {1} {4 \pi \varepsilon_0} \frac {Q_1 Q_2} {r^2} \vec r##

## The Attempt at a Solution

a) As the charges are equally spaced over that circumference, the net force acting over ##+Q## is zero, because the forcs diametrically opposed to each other cancel each other out.

b) If you take the "3'o'clock" particle out the circumference, the only particle that exerts force over ##+Q## is the particle diametrically oposed to "3'o'clock" particle: $$ \vec F= \frac {1} {4 \pi \varepsilon_0} \frac {q Q} {R^2} \vec R $$

c) The system back to equilibrium state.

What do you think guys? I'm not sure about my answers...