Chapter 2: Electrostatic Potential and Capacitance - Comprehensive NEET Physics Formulae Summary

1. Electrostatic Potential Energy

1.1 Formula: Electrostatic Potential Energy between Two Charges

Formula: $U = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r}$
Explanation: The electrostatic potential energy $U$ of a system of two point charges $q_{1}$ and $q_{2}$ separated by a distance $r$ . Here, $ϵ_{0}$ is the permittivity of free space.

1.2 Formula: Potential Energy in an External Field

Formula: $U = q V$
Explanation: The potential energy $U$ of a charge $q$ in an external electrostatic potential $V$ . This formula assumes the charge is placed in the external field without altering the field.

Derivations and Examples:

Derivation for Potential Energy between Two Charges: The formula is derived by considering the work done in bringing one charge from infinity to a point near another fixed charge.
Example Application: Calculate the potential energy between two charges, $3 \times 1 0^{- 6} C$ and $- 2 \times 1 0^{- 6} C$ , separated by 0.1 m.
Solution: $U = \frac{1}{4 π ϵ _{0}} \frac{( 3 \times 1 0 ^{- 6} ) ( - 2 \times 1 0 ^{- 6} )}{0.1} = - 0.54 J$

Common Mistake: Students often forget the negative sign when dealing with opposite charges, leading to incorrect answers.

2. Electrostatic Potential

2.1 Formula: Electrostatic Potential due to a Point Charge

Formula: $V = \frac{1}{4 π ϵ _{0}} \frac{Q}{r}$
Explanation: The electrostatic potential $V$ at a distance $r$ from a point charge $Q$ . This is a scalar quantity.

2.2 Formula: Potential Difference

Formula: $Δ V = V_{B} - V_{A} = \frac{1}{4 π ϵ _{0}} (\frac{Q}{r _{B}} - \frac{Q}{r _{A}})$
Explanation: The potential difference between two points A and B due to a charge $Q$ .

Derivations and Examples:

Derivation for Potential due to a Point Charge: Derived from the work done in bringing a unit positive test charge from infinity to a point in the field of charge $Q$ .
Example Application: Find the potential at a point 0.05 m away from a charge of $1 \times 1 0^{- 6} C$ .
Solution: $V = \frac{1 \times 1 0 ^{- 6}}{4 π \times 8.85 \times 1 0 ^{- 12} \times 0.05} = 179.8 kV$

NEET Tip: Remember that the potential is a scalar, so potentials due to multiple charges can be directly added algebraically.

3. Capacitance

3.1 Formula: Capacitance of a Parallel Plate Capacitor

Formula: $C = \frac{ϵ _{0} A}{d}$
Explanation: The capacitance $C$ of a parallel plate capacitor is proportional to the area $A$ of the plates and inversely proportional to the separation $d$ between them.

3.2 Formula: Energy Stored in a Capacitor

Formula: $U = \frac{1}{2} C V^{2}$
Explanation: The energy $U$ stored in a capacitor is a function of the capacitance $C$ and the potential difference $V$ across its plates.

Derivations and Examples:

Derivation of Capacitance Formula: Derived from the relationship between charge, electric field, and potential difference in a parallel plate capacitor.
Example Application: Calculate the capacitance of a capacitor with plates of area $2 m^{2}$ separated by $0.01 m$ in a vacuum.
Solution: $C = \frac{8.85 \times 1 0 ^{- 12} \times 2}{0.01} = 1.77 \times 1 0^{- 9} F$

Common Mistake: Ensure the unit of area is consistent (in square meters) and the separation is in meters.

4. Electric Field and Potential Relation

4.1 Formula: Relation between Electric Field and Potential Gradient

Formula: $E = - \frac{d V}{d r}$
Explanation: The electric field $E$ at a point is the negative gradient of the electric potential $V$ .

Derivations and Examples:

Derivation: The relation is derived by considering the work done by the electric field in moving a charge from one point to another.
Example Application: Given a potential function $V (x) = 5 x^{2}$ , find the electric field.
Solution: $E = - \frac{d ( 5 x ^{2} )}{d x} = - 10 x$

NEET Tip: Always remember that the electric field direction is from higher to lower potential.

Practice Questions

Calculate the potential at a point 0.1 m away from a charge of $5 \times 1 0^{- 6} C$ .
A parallel plate capacitor has plates of area $0.5 m^{2}$ and separation of $0.01 m$ . Find its capacitance.
Determine the energy stored in a capacitor of capacitance $2 μ F$ when connected to a 10 V battery.
If the potential function is given as $V (x) = 2 x$