### Coulombs Law

• Define charge and Coulomb's law.

• Explain the force of attraction between two charged bodies as a vector quantity.

### Solved Example:

#### 18-1-01

The electrical charge associated with one electron, in Coulombs, is:

Solution:
The elementary charge, usually denoted as e or sometimes q, is the electric charge carried by a single proton, or equivalently, the magnitude of the electric charge carried by a single electron, which has charge -e. This value is constant and equals to (-)1.6$\times 10^{-19}$ Coulombs.

### Solved Example:

#### 18-1-02

Two identical point charges with mass $m_1$ = $m_2$ = 0.60 grams are suspended in air with strings of length = 250mm. The total angle between them is 15$^\circ$. The charge each of them carrying is: Solution: Let tension in each string be T.
Equating horizontal components $T\sin 7.5^{\circ }=k\dfrac {q_{1}q_{2}}{r^{2}}$
Equating vertical components $T\cos 7.5^{\circ }=mg$
Dividing above two equations, $\tan 7.5^{\circ }=k\dfrac {q^{2}}{r^{2}mg}$ $q^{2}=\dfrac {r^{2}mg\tan 7.5^{\circ }}{k}$ where, $r=2\times 250\times \sin 7.5^{\circ } =65.26\ \mathrm{mm}$ $q=1.915\times 10^{-8}C=19.15\ \mathrm{nC}$

### Solved Example:

#### 18-1-03

Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 $\times$ 10$^{-6}$ C. The charges in the other two corners are -3.0 $\times$ 10$^{-6}$ C. What is the net force exerted on the charge in the top right corner by the other three charges?

Solution: $F = \dfrac{kq_1q_2}{r^2}$ \begin{align*} F_{14} = F_{12} &= \dfrac{(9\times 10^{9})(3\times 10^{-6})(3\times 10^{-6})}{(0.025)^2} = 129.5\ N \end{align*}

(F$_{12}$ towards q$_2$ and F$_{14}$ towards q$_4$)

\begin{align*} F_{13} &= \dfrac{(9\times 10^{9})(3\times 10^{-6})(3\times 10^{-6})}{(0.025\sqrt{2})^2}\\ &= 64.7\ N\ (\mathrm{away\ from}\ q_3\ \mathrm{as\ shown}) \end{align*} \begin{align*} \Sigma F_x &= -129.5 + 64.7 \cos 45^\circ = -83.75\ N \end{align*} \begin{align*} \Sigma F_y &= -129.5 + 64.7 \sin 45^\circ = -83.75\ N \end{align*} \begin{align*} F_R &= \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2}\\&= \sqrt{(83.75)^2 + (83.75)^2}\\&= 118.43\ N \end{align*}

### Solved Example:

#### 18-1-04

Three charges 4q, Q and q are in a straight line in the position of 0, $\dfrac{l}{2}$ and l respectively. The resultant force on q will be zero, if Q=? (RSMSSB Lab Asst. 2018) Solution: The force between 4q and q ${{F}_{1}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\cdot \dfrac{4q\times q}{{{l}^{2}}}$
The force between Q and q ${{F}_{2}}=\dfrac{1}{4\pi {{\varepsilon}_{0}}}\cdot \dfrac{Q\times q}{{{\left(\dfrac{l}{2}\right)}^{2}}}$
Since, the resultant force must be zero, \begin{align*} {{F}_{1}}+{{F}_{2}}&=0\\ \dfrac{4{{q}^{2}}}{{{l}^{2}}}&=-\dfrac{4Qq}{{{l}^{2}}}\\ Q&=-q \end{align*}

### Solved Example:

#### 18-1-05

Two charges of + 4 $\mu$C and -16 $\mu$C are separated from each other by a distance of 0.6 m. At what distance should a third charge of + 6 $\mu$C be placed from + 4 $\mu$C so that no force exerts on it will be zero? (DSSSB JE EE Oct 2019)

### Solved Example:

#### 18-1-06

The force between two charges $Q_1$ and $Q_2$ is given by: (NLC GET ECE Nov 2020)