Electrostatic Potential and Capacitance: Electric Potential Energy and Electric Potential - Lecture 02

Subject: Physics
Chapter: Electrostatic Potential and Capacitance
Lecture Number: 02
Target Exam: NEET 2027 (Lakshya Batch)
Instructor: Manish Raj (MR Sir)
Primary Topics Covered:
- Electric Potential Energy ( $U$ )
- Electric Potential ( $V$ )
- Relationship between Work, Potential Energy, and Potential

Work-Energy Theorem (WET):
- The total work done by all forces equals the change in kinetic energy ( $\Delta KE$ ).
- Equation: $\sum W = ΔKE$
- Breaking down the work components: $W_{CF} + W_{NCF} + W_{ext} + W_{int} + W_{psu} + W_{f} + W_{a} + ... = ΔKE$
- $W_{CF}$ : Work done by conservative forces.
- $W_{NCF}$ : Work done by non-conservative forces.
- $W_{ext}$ : Work done by external agents.
Relationship with Potential Energy ( $U$ ):
- Work done by a conservative force ( $W_{CF}$ ) is equal to searching for the negative change in potential energy: $W_{CF} = -ΔU$ .
- This relationship is always valid for conservative forces.
Conservation of Mechanical Energy (COME):
- Valid only when the work done by non-conservative forces is zero ( $W_{NCF} = 0$ ) and the work done by external agents is zero ( $W_{ext} = 0$ ).
- Formula: $(KE + U)_{i} = (KE + U)_{f}$ , where "i" stands for initial and "f" stands for final.

Fundamental Formula:
- For two point charges $q_1$ and $q_2$ separated by a distance $r$ : $U = \frac{k q_1 q_2}{r}$ .
System of Multiple Charges:
- The total potential energy of the system is the sum of the potential energies of every unique pair of charges.
- Number of Pairs: For a system of $N$ charges, the number of pairs is calculated as: $\frac{N(N - 1)}{2}$ .
- Example: 3-Charge System (Triangle):
- Charges $q_1, q_2, q_3$ at the vertices of a triangle.
- Total Pairs: $\frac{3(3-1)}{2} = 3$
- Total Potential Energy: $U = \frac{k q_1 q_2}{r_{12}} + \frac{k q_2 q_3}{r_{23}} + \frac{k q_1 q_3}{r_{13}}$ .
- Example: System of Identical Charges on a Square:
- 4 identical charges $q$ placed at the corners of a square with side length $L$ .
- Total Pairs: $\frac{4(4-1)}{2} = 6$
- Potential energy is calculated by summing the 4 side pairs and 2 diagonal pairs:
  - Side pairs: $4 \times \frac{k q^2}{L}$
  - Diagonal pairs: $2 \times \frac{k q^2}{\sqrt{2}L}$
  - U = \frac{4 k q^2}{L} + \frac{2 k q^2}{\sqrt{2}L} = \frac{k q^2}{L} [4 + \frac{2}{\sqrt{2}}] = \frac{k q^2}{L} [4 + \root{}{2}].

External Work ( $W_{ext}$ ):
- Assuming the process is done slowly (so $\Delta KE = 0$ ), the work done by an external agent is: $W_{ext} = ΔU = U_f - U_i$ .
Electric Field Work ( $W_{EF}$ ):
- The work done by the electric field (a conservative force) is: $W_{EF} = -ΔU = -(U_f - U_i) = U_i - U_f$ .
Calculation Example: Expanding a System:
- Problem: Find the work done by an external agent to increase the side of an equilateral triangle from $a$ to $2a$ with identical charges $q$ at the corners.
- Initial Energy ( $U_i$ ): $U_i = \frac{3 k q^2}{a}$ .
- Final Energy ( $U_f$ ): $U_f = \frac{3 k q^2}{2a}$ .
- $W_{ext} = U_f - U_i = \frac{3 k q^2}{2a} - \frac{3 k q^2}{a} = \frac{3 k q^2}{a} [\frac{1}{2} - 1] = -\frac{3 k q^2}{2a}$ .
- Problem: Work done by the electric field for the same expansion.
- $W_{EF} = U_i - U_f = \frac{3 k q^2}{a} - \frac{3 k q^2}{2a} = \frac{3 k q^2}{2a}$ .

Definition:
- Electric potential is the change in electric potential energy per unit charge.
- $\Delta V = \frac{ΔU}{q}$ .
- Alternatively, the work done in bringing a unit positive charge from one point to another.
Electrical Potential at a Point:
- The work done (slowly) in bringing a unit positive charge from infinity to a specific point in a field is called the electric potential ( $V$ ) at that point.
- Formula for a point charge $Q$ at distance $r$ : $V = \frac{k Q}{r}$ .
Properties:
- It is a scalar quantity (no direction).
- The unit is Joule per Coulomb ( $J/C$ ), which is defined as $1 \text{ volt}$ .
- It depends on the choice of reference point (commonly taken as $V = 0$ at infinity).
- When calculating, the charge $q$ must be inserted into formulas with its algebraic sign ( $+$ , $-$ ).

Conceptual Derivation:
- Starting from $W = ΔU$ and dividing by charge $q$ .
- Differential form: $dV = -\mathbf{E} ⋅ d\mathbf{r}$ .
- Integral form: $V_B - V_A = -∫_A^B \text{E} ⋅ d\text{r}$ .
Key Interpretations:
- Slope Concept: The electric field magnitude is the negative slope of the $V \text{ vs } r$ graph.
- $E = -\frac{dV}{dr}$ .
- Area Concept (The "Nali" Concept): The change in potential ( $\Delta V$ ) is the negative of the area under the $E \text{ vs } r$ graph.
- $Δ V = -∫ \text{E} ⋅ d\text{r}$ .

System of 5 Charges (Square and Centre):
- 4 charges $Q$ at corners, 1 charge $q$ at the center.
- Distance from corner to center is \frac{L}{\root{}{2}}.
- Potential energy ( $U_{system}$ ) requires summing all pairs (10 pairs total).
Bringing a charge from infinity to the center of a square:
- Initial potential ( $V_i$ ) at the center due to the 4 corner charges (assuming they are already there): V_c = 4 \times \frac{k Q}{(L / \root{}{2})} = \frac{4 \root{}{2} k Q}{L}.
- Work done to bring charge $q$ from infinity to center: W_{ext} = q \times (V_c - V_{∞}) = q V_c = \frac{4 \root{}{2} k Q q}{L}.

If potential $V$ is given as a function of coordinates $(x, y, z)$ , the electric field vector is found using partial differentiation:
- $\text{E} = - [ \frac{∂V}{∂x} \text{i} + \frac{∂V}{∂y} \text{j} + \frac{∂V}{∂z} \text{k} ]$

Homework: Perform a rough revision of today's lecture and the previous one.
Challenge Problem:
- Find the electric field at the center of a uniformly charged ring with linear charge density $\lambda$ where a small element of length $dl$ is cut from the ring.
- Hint: The field due to the remaining ring is equal and opposite to the field that would have been produced by the missing element $dl$ .
- $E = \frac{k dq}{R^2} = \frac{k (λ dl)}{R^2}$ .