Two charges $2 C$ and $6 C$ are separated by a finite distance. If a charge of $-4 C$ is added to each of them, the initial force of $12 \times 10^3 \,N$ will change to

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......$N$

The force between two point charges $A$ and $B$ is $F$. If $75\%$ of the charge of $A$ is transferred to $B$,then the new force between $A$ and $B$ is:

$5$ charges each of magnitude $10^{-5} \,C$ and mass $1 \,kg$ are placed (fixed) symmetrically about a movable central charge of magnitude $5 \times 10^{-5} \,C$ and mass $0.5 \,kg$ as shown in the figure. The charge at $P_1$ is removed. The acceleration of the central charge is [Given,$O P_1=O P_2=O P_3=O P_4=O P_5=1 \,m, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \,N \cdot m^2/C^2$]

$A$ force $F$ acts between sodium and chlorine ions of salt (sodium chloride) when placed $1 \ cm$ apart in air. The permittivity of air is $\varepsilon_0$ and the dielectric constant of water is $K$. When a piece of salt is placed in water,the electrical force acting between sodium and chlorine ions $1 \ cm$ apart is:

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 $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI Units}$)