A charge -q is placed at the center of a circle of radius r. Four charges -q, -q, +q, +q are placed at points A, B, C, D respectively on the circumference of the circle. Find the net electrostatic for

We have a charge $-q$ at the center O of a circle with radius $r$.
On the circumference, there are four charges:

• At A: $-q$
• At B: $-q$
• At C: $+q$
• At D: $+q$ The distance from each charge on the circumference to the center O is $r$.

The magnitude of the force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's Law: $F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2}$.
Let $F_0 = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$. All individual forces will have this magnitude.

• Force due to charge at A ($-q$) on charge at O ($-q$): Since both charges are negative, the force is repulsive. The charge at A is at the top, so it repels the charge at O downwards. $\vec{F}_{AO}$ is in the negative y-direction, with magnitude $F_0$.

• Force due to charge at B ($-q$) on charge at O ($-q$): Since both charges are negative, the force is repulsive. The charge at B is to the right, so it repels the charge at O to the left. $\vec{F}_{BO}$ is in the negative x-direction, with magnitude $F_0$.

• Force due to charge at C ($+q$) on charge at O ($-q$): Since the charges are opposite, the force is attractive. The charge at C is at the bottom, so it attracts the charge at O downwards. $\vec{F}_{CO}$ is in the negative y-direction, with magnitude $F_0$.

• Force due to charge at D ($+q$) on charge at O ($-q$): Since the charges are opposite, the force is attractive. The charge at D is to the left, so it attracts the charge