Lec 2.1: Electrostatics - Potential and Energy

The relationship between potential energy change and the electric field is defined by the work-charge relationship: $\Delta U = q \int \mathbf{E} \cdot d\mathbf{l}$ .
In a conservative electrostatic field, the integral over a closed loop is zero, often expressed as $\oint \mathbf{E} \cdot d\mathbf{l} = 0$ .
A mechanical analogy for potential energy is gravity, where $U = mgh$ . In electrostatics, potential energy is dependent on charge and position within an electric field.

Electric potential is the potential energy per unit charge. The unit of measure is the Volt ( $V$ ), which is equivalent to Joules per Coulomb ( $J/C$ ).
The relationship between potential difference and the electric field is: $\Delta V = - \int_i^f \mathbf{E} \cdot d\mathbf{l}$ .
For a point charge, the potential is given by $V = \frac{kq}{r}$ .
Integrals for potential always include a constant of integration ( $+C$ ).
By convention, potential is set to zero at infinity: $V = 0$ as $r \rightarrow \infty$ .
Calculating potential at a specific distance $a$ involves limits: $V_a = \int_a^{\infty} \mathbf{E} \cdot d\mathbf{l}$ .

The change in potential energy is proportional to the change in potential: $\Delta U = q \Delta V$ .
The unit for energy ( $U$ ) is Joules ( $J$ ).
Conservation of Energy principle states: $KE_i + U_i = KE_f + U_f$ .
To find the final velocity ( $v_f$ ) of a particle moving through a potential difference:
- $q(V_i - V_f) = \frac{1}{2} m v_f^2$
- $v_f = \sqrt{\frac{2q(V_i - V_f)}{m}}$

The total potential at a point due to multiple charges is the algebraic sum of the individual potentials (superposition principle): $V = \sum V_i = \sum \frac{k q_i}{r_i}$ .
For three charges, the potential at a point is: $V = V_1 + V_2 + V_3$ .

The total potential energy of a system ( $U_{sys}$ ) is the work required to assemble the charges from infinity.
This energy is calculated by summing the energy of every unique pair of charges (referred to as "bonds").
For a four-charge system ( $q_1, q_2, q_3, q_4$ ), the total energy is: $U_{sys} = U_{12} + U_{13} + U_{14} + U_{23} + U_{24} + U_{34}$ .
The individual energy between two charges is: $U_{ij} = \frac{k q_i q_j}{r_{ij}}$ .
It is incorrect to simply sum the total potential seen by each individual charge ( $U_1+U_2+U_3+U_4$ ) as that would double-count the interactions.

The force and potential energy are related by: $\Delta U = - \int \mathbf{F} \cdot d\mathbf{r}$ .
The electric field is the negative gradient of the potential: $\mathbf{E} = -\nabla V$ .
In spherical coordinates, the gradient is expressed as: $\nabla V = \frac{\partial V}{\partial r} \mathbf{\hat{r}} + \frac{1}{r} \frac{\partial V}{\partial \theta} \mathbf{\hat{\theta}} + \frac{1}{r \sin(\theta)} \frac{\partial V}{\partial \phi} \mathbf{\hat{\phi}}$ .
For the scope of this course, the angular components ( $\theta$ and $\phi$ ) are generally zero.
This simplifies the radial component of the electric field to: $E_r = -\frac{\partial V}{\partial r}$ .
Example validation for a point charge: If $V = \frac{kq}{r}$ , then $E_r = -\frac{d}{dr}(\frac{kq}{r}) = -kq(-r^{-2}) = \frac{kq}{r^2}$ .

To find the potential of a continuous distribution, sum the contributions of infinitesimal charge elements: $V = \int \frac{k \, dq}{r}$ .
For a ring of charge with total charge $Q$ and radius $R$ , at a point $x$ along the axis passing through the center:
- $V = \frac{kQ}{\sqrt{x^2 + R^2}}$
Gauss\'s Law link: $\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}$ . The electric field found via Gauss\'s Law can be integrated to find the potential difference: $\Delta V = -\int \mathbf{E} \cdot d\mathbf{r}$ .

Inside a conductor in electrostatic equilibrium, the electric field is zero: $E = 0$ .
Integrating the field from the surface ( $R$ ) to an internal point ( $r$ ): $\Delta V = -\int_R^r (0) \, dr = 0$ .
This implies that the potential is constant throughout the conductor and equal to its value on the surface: $V_{inside} = V_{surface} = \frac{kQ}{R}$ .
The potential calculation from infinity to the surface of a spherical conductor is: $V = - \int_{\infty}^{R} E \, dr = - \int_{\infty}^{R} \frac{kQ}{r^2} \, dr = \frac{kQ}{R}$ .

Scenario Setup:
- Charge 1 ( $q_1$ ): $4 \, \mu C$ located at $(0, 3 \, m)$ .
- Charge 2 ( $q_2$ ): $-6 \, \mu C$ located at $(4 \, m, 0)$ .
- Charge 3 ( $q_3$ ): $3 \, \mu C$ located at $(0, 0)$ .
- Point 4: Located at $(4 \, m, 3 \, m)$ .
Problem 1: Calculate the Potential ( $V$ ) at Point 4.
- $V_4 = V_1 + V_2 + V_3$
- Distance to $q_1$ from point 4 ( $r_{14}$ ) = $4 \, m$ .
- Distance to $q_2$ from point 4 ( $r_{24}$ ) = $3 \, m$ .
- Distance to $q_3$ from point 4 ( $r_{34}$ ) = $\sqrt{3^2 + 4^2} = 5 \, m$ .
- $V_4 = k \left( \frac{4 \times 10^{-6}}{4} + \frac{-6 \times 10^{-6}}{3} + \frac{3 \times 10^{-6}}{5} \right)$ .
Problem 2: Calculate the Potential Energy ( $U_3$ ) of Charge 3.
- $U_3 = q_3 V_{external} = q_3 (V_1 + V_2)$ .
- $U_3 = (3 \times 10^{-6} \, C) \times k \left( \frac{4 \times 10^{-6}}{3} + \frac{-6 \times 10^{-6}}{4} \right)$ .

In a uniform electric field ( $E$ ) between two parallel plates separated by distance $d$ , the potential difference is: $\Delta V = Ed$ .
The field ( $\mathbf{E}$ ) points from the positive plate to the negative plate.