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Electrostatic Potential and Capacitance

Physics • Chapter 2
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Electric Potential

Amount of work done in bringing a unit positive charge from infinity to a point.

$$ V = \frac{W}{q_0} $$

Unit: Volt (V) or J/C. It is a scalar quantity.

Electric Potential

Potential: Point Charge

Formula

$$ V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} $$

System of Charges

Total potential is the algebraic sum (Superposition Principle):

$$ V = V_1 + V_2 + \dots + V_n = \frac{1}{4\pi\varepsilon_0} \sum \frac{q_i}{r_i} $$

Potential: Dipole

At any point P at distance $r$ making angle $\theta$ with dipole moment $\vec{p}$:

$$ V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2} $$
Dipole Geometry
  • Axial ($\theta=0^\circ$): $V = \pm \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$
  • Equatorial ($\theta=90^\circ$): $V = 0$ (Everywhere on plane!)

Equipotential Surfaces

A surface with a constant value of potential at all points. Work done to move a charge on it is ZERO.

Equipotential Surfaces

Properties

  • Electric field is always perpendicular to it.
  • Two such surfaces never intersect.

Relation E & V

Potential Gradient

$$ E = -\frac{dV}{dr} $$

-ve sign shows Electric field points in the direction of decreasing potential.

Integral Form

$$ V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r} $$

Potential Energy ($U$)

System of Two Charges

$$ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} $$

Use signs of charges while calculating!

Charges in External Field

For single charge: $U = qV(\vec{r})$

For two charges:

$$ U = q_1V(\vec{r}_1) + q_2V(\vec{r}_2) + \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}} $$

Dipole P.E. in Field

Work done in rotating dipole in external field:

$$ U = -\vec{p} \cdot \vec{E} = -pE \cos\theta $$
Dipole in Field

Conductors & Shielding

  • Inside a conductor, $\vec{E} = 0$.
  • Surface of a conductor is an equipotential surface.
  • $\vec{E}$ at surface is perpendicular to it.
  • Interior has no excess charge.
Electrostatic Shielding

Electrostatic Shielding

Cavity inside conductor has zero field. Protects sensitive instruments.

Dielectrics & Polarisation

Non-conducting substances. In external $\vec{E}$, they develop induced dipole moment.

Polarisation

Polarisation ($P$): Dipole moment per unit volume. $P = \chi_e E$ ($\chi_e$ = electric susceptibility).

Capacitance ($C$)

Ability to store electric charge & energy.

$$ Q = CV \implies C = \frac{Q}{V} $$

Unit: Farad (F).

Isolated Spherical Conductor

$$ C = 4\pi\varepsilon_0 R $$

Shows $C$ depends only on geometry & medium.

Parallel Plate Capacitor

$$ C = \frac{\varepsilon_0 A}{d} $$

A = Area of plates, d = Distance between them.

Parallel Plate Capacitor

Electric Field Between Plates

$$ E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\varepsilon_0} $$

Effect of Dielectric

Inserting a dielectric slab (constant $K$) completely fills space:

$$ C = \frac{K\varepsilon_0 A}{d} = K \cdot C_{vacuum} $$
Dielectric Slab

Partial slab thickness $t$:

$$ C = \frac{\varepsilon_0 A}{d - t(1 - \frac{1}{K})} $$

Capacitors Combination

Series

$$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots $$

Charge $Q$ is same on all.

Parallel

$$ C_{eq} = C_1 + C_2 + \dots $$

Voltage $V$ is same across all.

Combinations

Energy Stored

$$ U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV $$

Energy Density $u = \frac{1}{2}\varepsilon_0 E^2$

Energy Loss

Sharing of Charge

Common Potential: $V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$

Energy Loss: $\Delta U = \frac{C_1C_2(V_1-V_2)^2}{2(C_1+C_2)}$