Electric charges of $1 \mu C, -1 \mu C$ and $2 \mu C$ are placed in air at the corners $A, B$ and $C$ respectively of an equilateral triangle $ABC$ having a side length of $10 \ cm$. The resultant force on the charge at $C$ is: (in $N$)

In a right-angled triangle $ABC$,$AB = 3 \ cm$,$BC = 4 \ cm$,and $\angle ABC = \frac{\pi}{2}$. Charges of $+15 \ e.s.u.$,$+12 \ e.s.u.$,and $-20 \ e.s.u.$ are placed at $A$,$B$,and $C$ respectively. What is the net force on the charge at $B$ in $dynes$?

Two identical conducting spheres,having charges of opposite sign,attract each other with a force of $0.108 \ N$ when separated by $0.5 \ m$. The spheres are connected by a conducting wire,which is then removed,and thereafter,they repel each other with a force of $0.036 \ N$. The initial charges on the spheres are

Three charges, each of $+1 \mu C$, are placed at the corners of an equilateral triangle. If the repulsive force between any two charges is $F$, then the net force on any one charge will be:

Two point charges $q_1 = 3 \mu \text{C}$ and $q_2 = -4 \mu \text{C}$ are placed at points $(2\hat{i} + 3\hat{j} + 3\hat{k})$ and $(\hat{i} + \hat{j} + \hat{k})$ respectively. Force on charge $q_2$ is . . . . . . $N$. (Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI Units}$)

Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. The charge on each is $6.5 \times 10^{-7} \; C$. Suppose the spheres $A$ and $B$ have identical sizes. $A$ third sphere of the same size but uncharged is brought in contact with the first,then brought in contact with the second,and finally removed from both. What is the new force of repulsion between $A$ and $B$?