Three equal charges of $4.0 * 10^{-7}\;C$ are located at the corners of a right triangle whose sides are $6 \;cm$, $8 \; cm$ and $10 \; cm$ respectively. Find the force exerted on the charge located at $90^{\circ}$ angle.

Given, (need figure)

Three equal charge ($q_1 = q_2 = q_3 = 4 * 10^{-7}\;C)$

Distance $(AB) = 8\;cm = 0.08\;m$

Distance $(BC) = 6\;cm = 0.06 \;m$

Distance $(AC) = 10\;cm = 0.1\;m$

Angle $ፈ ABC = 90 ^{\circ}$

Force $(F) = ?$

$F_{AB} = $$\frac{1}{4 \pi \epsilon_0}\frac{q_1\;*\;q_2}{R^2}$$ = 9 * 10^9 \;*\; $$\frac{(0.4 \;*\; 10^{-6})^2}{(0.08)^2}$$ = 0.225\;N$

$F_{BC} = $$\frac{1}{4 \pi \epsilon_0}\frac{q_1\;*\;q_2}{R^2}$$ = 9 * 10^9 \;*\; $$\frac{(0.4 \;*\; 10^{-6})^2}{(0.06)^2}$$ = 0.40\;N$

Force exerted on the charge located at $90 ^{\circ}$ is,

$F = \sqrt{(F_{AB})^2 + (F_{BC})^2 + 2\;F_{AB}\;F_{BC}\;Cos\theta}$

$F = \sqrt{(0.225)^2 + (0.4)^2 + 2\;*\;0.225\;*\;0.4\;*\;Cos\;90}$

⇒ $F = 0.46\;N$

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A solid sphere of radius $1\;cm$ is carrying a charge of $2\;C$. Find the electric field intensity at the center, on its surface and at a point $2\;cm$ from the center of the charged sphere.

Given,

Radius of the charged sphere $(R) = 1 \; cm = 0.01\;m$

Charge $(q) = 2\; C$

Electric field intensity $(E) = ?$

i) At the center of the sphere,

$E = 0$;          Because charge enclosed by Gaussian surface is zero. 

ii) At the surface of the sphere,

$E = $$\frac{1}{4 \pi \epsilon_0} \frac{q}{R^2}$$ \;=\; 9 \;*\; 10^9 \;*\; $$\frac{2}{(0.01)}$$ = 1.8 * 10^{12}\;N/C$

iii) At $2\;cm$ from the center,

i.e. $r = 2 = 2\;cm = 0.02\;m$

$E = $$\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$$ = 9 * 10^9 \;*\; $$\frac{2}{(0.02)}$$ = 9 \;*\; 10^{11}\; N/C$

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A hollow spherical conductor of radius 12 cm is charged to $6*10^{-6}C$. Find the electric field strength at the surface of sphere, inside the sphere at $8\;cm$ and at distance $15 \;cm$ from the sphere.

Given,

Radius of the charged sphere $(R) = 12 \; cm$

Charge $(q) = 6 * 10^{-6}\; C$

Electric field intensity $(E) = ?$

i) On the surface

$E = $$\frac{1}{4 \pi \epsilon_0} \frac{q}{R^2}$$ \;=\; 9 \;*\; 10^9 \;*\; $$\frac{6 \;*\; 10^{-6}}{(0.12)^2}$$ = 3.75 * 10^{6}\;N/C$

ii) Inside the sphere,

$E = 0$;          Because charge enclosed by Gaussian surface is zero.

iii) Outside the sphere at distance $15 \;cm$

i.e. $r = 12 + 15 = 27\;cm = 0.27\;m$

$E = $$\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$$ = 9 * 10^9 \;*\; $$\frac{6 \;*\; 10^6}{(0.27)^2}$$ = 7.41 \;*\; 10^5\; N/C$

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Two horizontal parallel plates each of area $500\;cm^2$ are mounted 2 mm apart in vacuum. The lower plate is earthed and the upper one is given a positive charge of $0.05\;\mu \; C$. Find the electric field intensity and the potential difference between the plates.

Given,

Area of the plates $(A) = 500 \; cm^2 = 500 * 10^{-4}\;m^2$

Distance $(d) = 2\;mm = 2 * 10^{-3}\;m$

charge $(q) = 0.05\;\mu C = 0.05 * 10^{-6}\;C$

Electric field intensity $(E) = \;?$

Electric Potential $(V) = \;?$

We have,

$E = $$\frac{\sigma}{\epsilon_0} = \frac{q}{\epsilon_0\;A} = \frac{0.05\; *\; 10^{-6}}{8.85 \;*\; 10^{-12} \;*\; 500 \;*\; 10^{-4}}$

        $= 112.9 * 10^3\; V/m$

And

$V = E \;*\; d$

    $ = 112.9 * 10^3 * 2 * 10^{-3} = 226\;V $

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An electron of mass $9.1 * 10^{-31}\;kg$ and charge $1.6 * 10^{-19}\;C$ is situated in a uniform electric field of intensity $1.2 * 10^{4}\;Vm^{-1}$. Find the time it takes to travel $1\; cm$ from rest.

Given,

Mass of electron $(m_e) = 9.1 \;* \;10^{-31}\;kg$

Charge of electron $(e) = 1.6 * 10^{-19}$

Electric field intensity $(E) = 1.2 * 10^4\;V/m$

Distance covered $(s) = 1\;cm = 100^{-2}\;m$

Time taken (t) = ?

We have,

$F = eE$

or, ma = eE

or, $a = $$\frac{eE}{m} = \frac{1.6 * 10^{-19} \;*\; 1.2 \;*\; 10^4}{9.1 \;*\; 10^{-31}}$$ = 2.1 * 10^{15}\;m/s^2$

Now we have to calculate,

$S = ut \;+\; $$\frac{1}{2}$$at^2$

$S = 0 \;+\; $$\frac{1}{2}$$at^2$                                              [∵ $u = 0 \; m/s$]

$0.01 = $$\frac{1}{2}$$ \;*\; 2.1 \;*\; 10^{15} \;*\; t^2$

⇒ $t = 3.1 \;*\; 10^{-9}\;Sec $

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An alpha particle is a nucleus of doubly ionized helium. It has mass of $6.68 * 10^{-27}$kg and charge of $3.2 * 10^{-19}C$. Compare the force of electrostatic repulsion between the two alpha particles with the force of gravitational attraction between them.

Given,

Mass of the alpha particle = $6.68\;*\;10^{-27}\;kg$

Charge of the alpha particle = $3.2 * 10^{-19}\;C$

We have,

Electrostatic Force $(E_f) = $$\frac{1}{4\pi \epsilon_0} \frac{q_1\;q_2}{R^2}$

Electrostatic Force $(E_f) = 9*10^9 * $$\frac{q^2}{R^2}$ = $9*10^9 * $$\frac{(3.2 * 10^{-19})^2}{R^2}$ ......... (i)

Gravitational force $(G_f) = G $$\frac{M_1\;M_2}{R^2}$

Gravitational force $(G_f) = 6.67\;*\;10^{-11} $$\frac{(6.68*10^{-27})^2}{R^2}$ .......... (ii)

Dividing equation (i) and (ii), we get

$\frac{E_f}{G_f} = \frac{9.216 \;*\; 10^{-28}}{2.976 \;*\; 10^{-63}}$

$= 3.096 \;*\; 10^{35}$

⇒ $E_f = 3.096 \;*\; 10^{35}\; G_f$

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E4.1 Coulomb's Law:

A French Scientist, Coulomb (In 1875): First experimentally measured the force of attraction (or repulsion) between two charges. On the basis of his measurements, Coulomb enunciated a law, called Coulomb's Law.  

"Two charges attract  (or repel) each other with the force, which is directly proportional to the product of the magnitude of the charge and inversely proportional to the square of the distance between them." 

The force is along the straight line joining them, as shown in figure below. This law is based on the inverse square law.

Let $Q_1$ and $Q_2$ be the two charges, separated by a distance $(r)$ then,

 Mathematically,

$F = k.\frac{Q_1 Q_2}{r^2}$ .......... (i)

Where, $k$ is the constant of proportionality. It's value depends upon the medium between the charges. If the charges are situated in vacuum (or air) in SI Unit,

$F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2}$ .......... (ii)

Where, $\epsilon_0$ = permittivity of free space (or vacuum) = $8.854 * 10^{-12} \frac{C^2}{N\;m^2}$.

If the two charges have the same sign, the electrostatic force between them is repulsive. If they have different signs, the electrostatic force between them is attractive.

Here is the video about Coulomb's Law.

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