Write some important points for vector form of Coulomb’s law.
$\overrightarrow{\mathrm{F}_{21}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{21}^{2}} \cdot \hat{r}_{21}$
and equation is true for both positive and negative values of $q_{1}$ and $q_{2}$.
If $q_{1}$ and $q_{2}$ both are positive or both are negative, then $\overrightarrow{F_{21}}$ is in direction of $\overrightarrow{r_{21}}$ which shows repulsion (like charges).
If $q_{1}$ and $q_{2}$ are unlike charges, then $\overrightarrow{F_{21}}$ is along $\hat{r}_{21}\left(=-\hat{r}_{12}\right)$ which shows attraction.
By replacing 1 and 2 in above equations $(1)$, $\overrightarrow{\mathrm{F}_{12}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r_{12}^{2}} \hat{r}_{12}=-\overrightarrow{\mathrm{F}}_{21}$
Coulomb's law is agrees with Newton's third law.
If two charges are placed in medium, then the force between them becomes $\frac{1}{K}$ times means Coulomb force decreases.
Coulomb forces are central forces means they pass through line connecting centres of charges. Coulomb's law is inverse square law.
According to this law, electric force are of two types : attractive and repulsive.
There is no effect of third charge on electric force between two charges. Thus, Coulomb force is called two body 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? $\left(m_{p}=1.67 \times 10^{-27} \,kg , m_{e}=9.11 \times 10^{-31}\, kg \right)$
A particle of mass $1 \,{mg}$ and charge $q$ is lying at the mid-point of two stationary particles kept at a distance $'2 \,{m}^{\prime}$ when each is carrying same charge $'q'.$ If the free charged particle is displaced from its equilibrium position through distance $'x'$ $(x\,< \,1\, {m})$. The particle executes $SHM.$ Its angular frequency of oscillation will be $....\,\times 10^{8}\, {rad} / {s}$ if ${q}^{2}=10\, {C}^{2}$
Point charge $q$ moves from point $P$ to point $S$ along the path $PQRS$ (figure shown) in a uniform electric field $E$ pointing coparallel to the positive direction of the $X - $axis. The coordinates of the points $P,\,Q,\,R$ and $S$ are $(a,\,b,\,0),\;(2a,\,0,\,0),\;(a,\, - b,\,0)$ and $(0,\,0,\,0)$ respectively. The work done by the field in the above process is given by the expression
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
Two charged spheres separated at a distance $d$ exert a force $F$ on each other. If they are immersed in a liquid of dielectric constant $2$, then what is the force (if all conditions are same)