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Chapter 1: Electric Charges and Fields – Formulas (MathJax)

Chapter 1: Electric Charges and Fields

This section provides a quick reference to the essential formulas from the chapter “Electric Charges and Fields.” Understanding these formulas is key to solving numerical problems.

1. Coulomb’s Law

Describes the electrostatic force between two point charges.

\[ F = k \frac{|q_1 q_2|}{r^2} \]
Where:
  • \( F \) = electrostatic force
  • \( k \) = Coulomb’s constant (\( 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2} \))
  • \( q_1, q_2 \) = magnitudes of the point charges
  • \( r \) = distance between the charges
\[ k = \frac{1}{4\pi\epsilon_0} \]
Where:
  • \( \epsilon_0 \) = permittivity of free space (\( 8.854 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} \))

In vector form:

\[ \vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21} \]
Force on \( q_1 \) due to \( q_2 \).

2. Electric Field

The electric field at a point is the force experienced by a unit positive test charge placed at that point.

\[ \vec{E} = \frac{\vec{F}}{q_0} \]
Where:
  • \( \vec{E} \) = electric field vector
  • \( \vec{F} \) = force on test charge \( q_0 \)
  • \( q_0 \) = test charge

Electric field due to a point charge Q:

\[ E = \frac{1}{4\pi\epsilon_0} \frac{|Q|}{r^2} \]
Magnitude of electric field at distance \( r \) from charge \( Q \).

3. Electric Dipole

A pair of equal and opposite charges (\( +q \) and \( -q \)) separated by a small distance \( 2a \).

\[ \text{Electric Dipole Moment } (\vec{p}) = q (2\vec{a}) \]
Direction is from \( -q \) to \( +q \).

Electric field on the axial line (at distance \( r \) from center):

\[ \vec{E}_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}r}{(r^2 – a^2)^2} \]
For \( r \gg a \), \( \vec{E}_{\text{axial}} \approx \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3} \)

Electric field on the equatorial plane (at distance \( r \) from center):

\[ \vec{E}_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{-\vec{p}}{(r^2 + a^2)^{3/2}} \]
For \( r \gg a \), \( \vec{E}_{\text{equatorial}} \approx \frac{1}{4\pi\epsilon_0} \frac{-\vec{p}}{r^3} \)

Torque on a dipole in a uniform electric field:

\[ \vec{\tau} = \vec{p} \times \vec{E} \]
Magnitude: \( \tau = pE \sin\theta \)

4. Electric Flux

A measure of the number of electric field lines passing through a surface.

\[ \Phi_E = \int \vec{E} \cdot d\vec{A} \]
For a uniform field and flat surface: \( \Phi_E = EA \cos\theta \)

5. Gauss’s Law

Relates the electric flux through any closed surface to the net electric charge enclosed within that surface.

\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0} \]
Where \( q_{\text{enclosed}} \) is the total charge inside the closed surface.

6. Applications of Gauss’s Law (Electric Fields)

(i) Infinite Long Straight Uniformly Charged Wire:

\[ E = \frac{\lambda}{2\pi\epsilon_0 r} \]
Where \( \lambda \) = linear charge density.

(ii) Uniformly Charged Infinite Plane Sheet:

\[ E = \frac{\sigma}{2\epsilon_0} \]
Where \( \sigma \) = surface charge density.

(iii) Uniformly Charged Thin Spherical Shell (Radius R):

\[ E_{\text{out}} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \quad (\text{for } r \ge R) \] \[ E_{\text{in}} = 0 \quad (\text{for } r < R) \]
\( Q \) is the total charge on the shell.

(iv) Uniformly Charged Solid Sphere (Radius R):

\[ E_{\text{out}} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \quad (\text{for } r \ge R) \] \[ E_{\text{in}} = \frac{1}{4\pi\epsilon_0} \frac{Qr}{R^3} \quad (\text{for } r < R) \]
\( Q \) is the total charge on the sphere.

Remember to practice applying these formulas with numerical problems to solidify your understanding.

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