Two charges ${q_1}$ and ${q_2}$ are placed in vacuum at a distance $d$ and the force acting between them is $F$. If a medium of dielectric constant $4$ is introduced around them, the force now will be

A point charge of $40$ stat coulomb is placed $2$ $cm$ in front of an earthed metallic plane plate of large size. Then the force of attraction on the point charge is.....$dynes$

Two charges each equal to $2\,\mu C$ are $0.5\,m$ apart. If both of them exist inside vacuum, then the force between them is.......$N$

A proton is fired at an initial velocity of $150 \,m/s$ at an angle of $60^o $ above the horizontal into a uniform electric field of $2 \times 10^{-4} \,N/C$ between two charged parallel plates as shown in figure. Then the total time the particle is in motion 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 given below. The charge at $P_1$ is removed. The acceleration of the central charge is [Given, $\left.O P_1=O P_2=O P_3=O P_4=O P_5=1 m , \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right]$

If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?