Why Coulombian force is called two body force ?
Three charges are placed as shown in figure. The magnitude of $q_1$ is $2.00\, \mu C$, but its sign and the value of the charge $q_2$ are not known. Charge $q_3$ is $+4.00\, \mu C$, and the net force on $q_3$ is entirely in the negative $x-$ direction. As per the condition given the sign of $q_1$ and $q_2$ will be
The radius of two metallic spheres $A$ and $B$ are ${r_1}$ and ${r_2}$ respectively $({r_1} > {r_2})$. They are connected by a thin wire and the system is given a certain charge. The charge will be greater
Three point charges $q,-2 q$ and $2 q$ are placed on $x$-axis at a distance $x=0, x=\frac{3}{4} R$ and $x=R$ respectively from origin as shown. If $q =2 \times 10^{-6}\,C$ and $R =2\,cm$, the magnitude of net force experienced by the charge $-2 q$ is .......... $N$
Two identical spheres each of radius $R$ are kept at center-to-center spacing $4R$ as shown in the figure. They are charged and the electrostatic force of interaction between them is first calculated assuming them point like charges at their centers and the force is also measured experimentally. The calculated and measured forces are denoted by $F_c$ and $F_m$ respectively.
($F_c$ and $F_m$ denote magnitude of force)
Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.
$(a)$ Compare the strength of these forces by determining the ratio of their magnitudes $(i)$ for an electron and a proton and $(ii)$ for two protons.
$(b)$ Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \mathring A \left( { = {{10}^{ - 10}}m} \right)$ apart? $\left(m_{p}=1.67 \times 10^{-27} \,kg , m_{e}=9.11 \times 10^{-31}\, kg \right)$