In a hydrogen atom,the distance between the electron and proton is $2.5 \times 10^{-11} \ m$. The electrical force of attraction between them will be

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

$5$ 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,$O P_1=O P_2=O P_3=O P_4=O P_5=1 \,m, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \,N \cdot m^2/C^2$]

The magnitude of the electric force on a $2\, \mu C$ charge placed at the centre $O$ of two equilateral triangles,each of side $10\, cm$,as shown in the figure,is $P$. If the charges at $A, B, C, D, E,$ and $F$ are $2\, \mu C, 2\, \mu C, 2\, \mu C, -2\, \mu C, -2\, \mu C,$ and $-2\, \mu C$ respectively,then $P$ is: ..... $N$

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

The sum of two point positive charges separated by a distance of $1.5 \ m$ in air is $25 \mu C$. If the electrostatic force between the two charges is $0.6 \ N$,then the difference between the two charges is (in $\mu C$)