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# Coulomb's Law - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Coulomb's Law is considered one the most difficult concept.

• 69 Questions around this concept.

## Solve by difficulty

Two identical charged spheres suspended from a common point by two massless strings of length $\dpi{100} l$ are initially a distance $\dpi{100} d(d< < l)$ apart because of their mutual repulsion. The charge begins to leak from both spheres at a constant rate. As a result, the charges approach each other with a velocity $\dpi{100} \upsilon$. Then the relation between v and x is:

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q/q equals

Two spherical conductors B  and C   having equal radii and carrying equal charges in them repel each other with a force  F   when kept apart at some distance.A   third spherical conductor having same radius as that of  B   but uncharged is brought in contact with  B   then brought in contact with   C  and  finally removed away from both, The new force of repulsion between  B  and  C   is  :

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 pendulum bob of mass and carrying a charge is at rest in a horizontal uniform electric field of 20000 V/m. The tension in the thread of the pendulum is ( g = 10 m/s2) :

In the given figure two tiny conducting balls of identical mass m and identical charge $q$ hang from non-conducting threads of equal length L. Assume that $\tan\theta \approx \sin\theta$ is so small that, then for equilibrium $x$ is equal to

## Concepts Covered - 1

Coulomb's Law

Coulomb's Law: The force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

$\\F\propto \frac{Q_{1}Q_{2}}{r^{2}} \\ \\ \\F=\frac{KQ_{1}Q_{2}}{r^{2}}$

K = Proportionality Constant

$Q_1$ and $Q_2$ are two Point charges

In SI unit value of K is

$K=\frac{1}{4 \pi \varepsilon_{0}}$

Where,

$\left(\varepsilon_{0}\right)=8.85 \times 10^{-12} \frac{C^{2}}{N-m^{2}}$  known as absolute permittivity of air or free space

The vector form of Coulomb's Law:

Consider two charges $q_1$ and $q_2$ separated by a distance r. Let the position vectors of  $q_1$ be $r_1$ and that of $q_2$ be $r_2$. Then the force due to $q_2$ on $q_1$ as shown in figure$F_{12}$ is directed along the unit vector $r_{12}$ and

$\\ \ \ \ \ \ \ \ \ \ \ \ F_{12}=\frac{Kq_1q_2}{r^2}\hat r_{12}\\ \\ \\ here, \ \hat r_{12}=\frac{\vec r_1-\vec r_2}{|r_1-r_2|}=\frac{\vec r_{12}}{r}\\ \\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F_{12}=\frac{Kq_1.q_2}{r^3}\vec r_{12}$

Force when dielectric inserted between the charges:

When a dielectric of dielectric constant k is completely filled between the charges then force

$F_{med}=\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}kr^{2}}=\frac{q_{1}q_{2}}{4\pi \varepsilon _{0}\epsilon_rr^{2}}$

$\epsilon_r$ is relative permittivity / dielectric constant of the medium. The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space. (dielectric will be explained later in detail in this chapter)

If the dielectric of thickness d is partially filled between the charges $Q_1 \ \ and \ \ Q_2$  then

$F=\frac{Q_1Q_2}{4\pi \epsilon _0(r-d+\sqrt{k}d)^2}$

Principle of Superposition:

It states that the total force acting on a given charge due to a number of charges is the Vector sum of the individual forces acting on that charge due to all the charges.

## Study it with Videos

Coulomb's Law

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