Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is
$F / 8$
$F / 4$
$4 F$
$F / 16$
$5$ 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 given below. The charge at $P_1$ is removed. The acceleration of the central charge is [Given, $\left.O P_1=O P_2=O P_3=O P_4=O P_5=1 m , \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right]$
The value of electric permittivity of free space is
When the distance between the charged particles is halved, the force between them becomes
Write general equation of Coulombian force on ${q_1}$ by system of charges ${q_1},{q_2},.......,{q_n}$.
Similar charges are placed at corners of a square and a charge $q_0$ is placed at it's centre find net force on it