Electric Charges And Fields

Electric Charges and Fields is a key chapter in electromagnetism that explains the basics of electric charge—how objects can be positively or negatively charged, and how like charges repel while opposites attract. It covers important properties like charge conservation and quantization. The chapter introduces Coulomb’s Law to measure the force between charges, and explains electric fields—invisible regions around a charge where other charges feel a force. It also discusses electric field lines, dipole behavior in fields, and Gauss’s Law, which helps calculate electric fields in cases with symmetry.

1.0Introduction to Electric Charges

Electric charge: It’s a fundamental property of matter responsible for electric and magnetic interactions.

Types of Charges

1. Positive charge – Occurs when there is a loss of electrons.
2. Negative charge – Occurs when there is a gain of electrons.

Units of Charge

• SI - Coulomb(C)
• CGS-Stat coulomb (stat C) 
• Dimensional Formula -[AT]

2.0Properties of Electric Charges

1. Additivity of electric Charge.
2. Conservation of electric Charge
3. Quantization of Electric Charge
4. Charge can be transferred
5. Charge is invariant.
6. Charge is a scalar quantity
7. Charge is always associated with mass

3.0Coulomb’s Law

The electrostatic force between two point charges is directly proportional to their charges and inversely proportional to the square of the distance between them, always acting along the line joining them.

$F=K\frac{q_1{q}_2}{r^2}$

$k=\frac{1}{4\pi {ϵ}_0}=9\times 10^{9}N{m}^2{C}^{-2}$=Electrostatic constant or Coulomb’s Constant

Coulomb's Law in Vector Form

${\vec{F}}_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2\left({\vec{r}}_2-{\vec{r}}_1\right)}{|{\vec{r}}_2-{\vec{r}}_1{|}^3}\text{or}{\vec{F}}_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2\left({\vec{r}}_1-{\vec{r}}_2\right)}{|{\vec{r}}_1-{\vec{r}}_2{|}^3}$

4.0Superposition Principle

When multiple charges are present, the net force on any charge is the vector sum of forces from all the other charges.

$\vec{F}={\vec{F}}_{01}+{\vec{F}}_{02}+{\vec{F}}_{03}+\dots +{\vec{F}}_{0n}$

$\vec{F}=\frac{k{q}_0{q}_1}{r_1^2}{\hat{r}}_1+\frac{k{q}_0{q}_2}{r_2^2}{\hat{r}}_2+\dots +\frac{k{q}_0{q}_n}{r_n^2}{\hat{r}}_n$

Relative Permittivity(${ϵ}_r$) or Dielectric Constant (K)

$\text{Dielectric Constant (K)}=\frac{\text{Permittivity of a Medium }(\epsilon )}{\text{Permittivity of Vacuum }({\epsilon }_0)}$

$K=\frac{{\epsilon }_r}{{\epsilon }_0}$

$\epsilon ={\epsilon }_0{\epsilon }_r$

The value of ${\epsilon }_r$ or K ≥1

5.0Electric Field 

• An electric field is the region around a charge where it exerts a force on other charges.
• Mathematically, $\vec{E}=\frac{\vec{F}}{q_0}$ SI unit : N/C or V/m
• $\vec{E}=\underset{q_0\to 0}{\lim }\frac{\vec{F}}{q_0}$
• Dimensional Formula  $[M^1{L}^1{T}^{-3}{A}^{-1}]$
• It is a vector quantity.

6.0Electric Field Lines

• Electric field lines are imaginary lines, straight or curved, that represent the direction and strength of the electric field around a charged object. At any point on these lines, the tangent indicates the direction of the electric field at that location.
• Two electric field lines can never intersect each other.

• Electrostatic field lines can never form closed loops.

Electric field lines due to positive or negative charges

Electric line of force due to an electric dipole

7.0Continuous Charge Distribution

A group of closely spaced electric charges forms a continuous charge distribution.

8.0Electric Dipole

• It is a pair of equal and opposite charges separated by a small distance.

• The dipole moment is the product of the magnitude of either charge and the distance between them.$\vec{p}=q(\overrightarrow{2l})$
• S.I. unit- Cm

Electric Field Due to a Dipole

1. At Axial / End on position

$E_{Axial}=\frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}$

2. At Equator/Broadside on position

$E_{Equitorial}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^3}$

3. At general position

$E=\frac{p}{4\pi {\epsilon }_0{r}^3}\sqrt{3co{s}^2\theta +1}$

$Tan\alpha =\frac{1}{2}Tan\theta$

Electric Field Intensity due to a charged wire

Special Cases:

1. For Infinite Wire,( both ends goes to infinite)

         $EX=E\perp =\frac{2K\lambda }{r}$       

         E_Y=E॥=0

2. For Semi-infinite wire

EX=E丄=$\frac{K\lambda }{r}$

$E_Y=E_{||}=\frac{K\lambda }{r}$

$E_{Resultant}=\sqrt{2}\frac{K\lambda }{r}$

3. Electric field due to finite wire at symmetric point:

$E_X=E_{\perp }=\frac{2K\lambda }{r}sin\theta$

$E_Y=E_{॥}=0$

4. Electric Field due to a uniformly charged Arc

$E=\frac{2K\lambda }{R}Sin(\frac{\theta }{2})$

5. Electric field at centre of uniformly charged Ring

$E=\frac{2K\lambda }{R}Sin(\frac{\theta }{2}),\theta \to$ angle of arc

$\theta$ =360⁰

$E=\frac{2K\lambda }{R}Sin(\frac{360}{2})$

E =0

6. Electric Field due to a Uniformly charged Ring at its Axis

$E=\frac{kQx}{(R^2+x^2{)}^{3/2}}$

Special cases:

1. Electric field on the axis for small values of x : $E=\frac{kQ}{R^3}\cdot x$
2. Electric field at the centre of the ring is zero because x=0
3. Electric field at the axis for larger values of x: $E\approx \frac{kQ}{x^2}$
4. Maximum value of electric field

$\frac{dE}{dx}=0$

$x=\pm \frac{R}{\sqrt{2}}$

$E_{\text{max}}=\frac{2kQ}{3\sqrt{3}{R}^2}$

9.0Dipole placed in an electric field, torque acts on it

て=p E Sin  

(∴ $\theta$is the angle between dipole moment (p)and electric field (E))

$\vec{\tau }=\vec{p}\times \vec{E}$

Special Cases:

• If =0⁰ then stable Equilibrium.
• If =180⁰ then て=o, unstable equilibrium.

10.0Work done in rotating a dipole in a  uniform electric field

When an electric dipole with dipole moment is oriented at an angle to an electric field , torque is exerted on the dipole, causing it to rotate. The resulting work done can be expressed as

$W=pE(Cos{\theta }_1-Cos{\theta }_2)$

11.0Electric Flux

• This physical quantity is used to measure strength of electric field and it is defined as the total number of electric field lines passing through an area.
• Electric flux is a scalar quantity.
• Unit of Electric Flux:Nm2/C or V m
• Dimensional Formula-[M¹L³T^-3A^-1]
• Electric flux through a large surface$\varphi =\int d\varphi =\int \vec{E}\cdot d\vec{A}$
• If  Electric Field is uniform=$\Rightarrow \vec{E}$=Constant(same everywhere)

$\varphi =\vec{E}\cdot \int \overrightarrow{dA}(\because \int \overrightarrow{dA}=\text{total area vector of a plane surface})$

$\Rightarrow \varphi =\vec{E}\cdot \vec{A}\Rightarrow \varphi =EA\cos \theta$

Different cases

1. If $\theta$=0⁰, $\varphi =EA\cos 0^{\circ }=EA\text{(positive flux means outgoing or leaving)}$

2. If $\theta$ is 90⁰, $\varphi =EAcos90°=0.$
3. If $\theta$ is 180⁰ , $\varphi =EACos180°=-EA$ (negative flux means incoming or entering)

12.0Gauss’s Law

According to this law the total electric flux ($\varphi$) through any closed surface (S) in free space is equal to $\frac{1}{{\epsilon }_0}$ times the total electric charge (q) enclosed by the surface.

$\varphi =\oint \vec{E}.\overrightarrow{dS}=\frac{Q_{enclosed}}{{\epsilon }_0}$

13.0Applications of Gauss Law

1. Electric field Intensity due to infinitely long wire

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2K\lambda }{r}$

2. Electric field due to uniformly charged long cylindrical pipe/cylindrical shell

• Case 1.Electric field at any point outside the cylinder(r>R): $E=\frac{\sigma R}{{ϵ}_{0}r}$
• Case 2.For the point lying on the surface(r≈R): $E=\frac{\sigma }{{\epsilon }_0}$
• Case 3.For the point inside the surface(r<R): $E_{inside}=0$

3. Electric Field due to Uniformly Charged Infinite Sheet

• Non Conducting Sheet: $E=\frac{\sigma }{2{\epsilon }_0}$
• Conducting sheet or Metal Plate: $E=\frac{\sigma }{{\epsilon }_0}$

4. Electric Field due to the charged conducting sphere or charged thin shell

• Electric Field at any point outside the sphere (r>R)

$E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{kq}{r^2}$

• For any point lying on the surface of sphere (r=R)

$E_s=\frac{kq}{R^2}=\frac{q}{4\pi {\epsilon }_0{R}^2}=\frac{\sigma }{{\epsilon }_0}$        $(\because \sigma =\frac{q}{4\pi R^2})$

• Electric Field at any point Inside the sphere(r<R)

In this case charge enclosed by the gaussian surface is zero, i.e., E_inside=0

• Variation of E with r 

4. Electric Field due to uniformly charged non conducting sphere(solid sphere)

• Electric field at any point outside the sphere (r>R)

$E=\frac{kq}{r^2}$

• Electric field at any point lying on the surface of sphere(r=R)

$E_S=\frac{\rho R}{3{\epsilon }_0}$     (∴ $\rho$ is volume charge density)

• Electric field at any point inside the sphere(r<R)

$E_{inside}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^3}r=\frac{rho}{3{\epsilon }_0}(r)$

• Variation of E with r

14.0Sample Questions on Electric Charges And Fields

Q-1. $10^{10}$ alpha particles are ejected per second from a body, then after how much time, the body will acquire a charge of 8 μC ?

Solution: Charge appear per second on body, $Q=(\frac{N}{t})q_{\alpha }$

$Q=10^{10}\times (2\times 1.6\times 10^{-19})=3.2\times 10^{-9}C/sec$

Let  t  be the time taken to acquire charge of 8 μC then

$t=\frac{8\times 10^{-6}}{3.2\times 10^{-9}}=2.5\times 10^{3}sec$

Q-2.Find final charges on the spheres when switch S is closed.

Solution: Total charge of the system = $Q=60\mu C+0=60\mu C$

Then after conduction

$\frac{Q_1^{\prime }}{Q_2^{\prime }}=\frac{R_1}{R_2}\Rightarrow \frac{2}{3}$

$Q{1}^{\prime }=(\frac{2}{2+3})60\mu C=24\mu C$

$Q{2}^{\prime }=(\frac{3}{2+3})60\mu C=36\mu C$

Q-3.Why can we not use Coulomb’s law for large size bodies ?

Solution:

When large size charged conducting spheres brought close to each other, there charges moves away due to repulsion hence effective distance between their centers increases r'>r

$F_{actual}=F_{calculated}=k\frac{q_1{q}_2}{r^2}$

Q-4.Two identical conducting spheres A and B carry equal charges and are placed at a distance r apart in vacuum. The electrostatic force between them is F. A third identical uncharged sphere C is first brought into contact with sphere A, then with sphere B, and is finally removed. What will be the new electrostatic force between spheres A and B?

Solution: 

$F=k\frac{q^2}{r^2}$……….(1)

Now C (uncharged sphere) is touched with A first

Then C is touched with B

New force between A & B now is

$F^{\prime }=\frac{k\left(\frac{Q}{2}\right)\left(\frac{3Q}{4}\right)}{r^2}=\frac{3k{Q}^2}{8r^2}=\frac{3F}{8}$

Q-5.Two equally charged spheres are placed far apart. If the gravitational force equals the electrostatic force between them, find the ratio of specific charge $\frac{q}{m}$.

Solution: Equating the gravitational and electrostatics force 

$\frac{k{q}^2}{r^2}=\frac{G{m}^2}{r^2}\Rightarrow \frac{q}{m}=\sqrt{\frac{G}{k}}$$=\sqrt{\frac{6.67\times 10^{-11}}{9\times 10^9}}=\sqrt{\frac{0.74}{10^{20}}}=\frac{0.86}{10^{10}}=0.86\times 10^{-10}$