Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly and without acceleration. It is a scalar quantity. $V = \dfrac{W_\infty}{q_0}$
Fig 1.1 — Point charge +q, test point P at distance r; dashed circles are equipotential surfaces
Derivation
Consider point charge $+q$ at origin. Bring test charge $q_0$ from $\infty$ to point P at distance $r$.
①
At distance $x$ from $q$, electric force on $q_0$ is directed away (repulsive).
The external agent must apply equal and opposite force:
$$F_{ext} = \frac{kq\,q_0}{x^2} \quad \text{(directed toward }q\text{)}$$
②
Small work done by external agent in displacement $dx$ (from $\infty$ toward $q$, so $dx$ is negative):
$$dW = F_{ext}\cdot(-dx) = -\frac{kq\,q_0}{x^2}\,dx$$
③
Total work done from $\infty$ to $r$:
$$W = \int_{\infty}^{r} \left(-\frac{kq\,q_0}{x^2}\right)dx
= -kq\,q_0\left[-\frac{1}{x}\right]_{\infty}^{r}
= \frac{kq\,q_0}{r}$$
④
Electric potential $V = \dfrac{W}{q_0}$:
★ Result
$$V = \frac{W}{q_0} = \frac{kq}{r} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q}{r}$$
An electric dipole has charges $+q$ and $-q$ separated by $2a$. Dipole moment $p = q \cdot 2a$, directed from $-q$ to $+q$. Potential at any point is the scalar sum of potentials due to each charge.
Fig 2.1 — Point P at $(r, \theta)$; $r_1$ = dist from $-q$, $r_2$ = dist from $+q$
Derivation
For $r \gg a$, using geometry: $r_1 \approx r + a\cos\theta$ and $r_2 \approx r - a\cos\theta$.
①
Potential at P due to $+q$ and $-q$ (scalar sum):
$$V = \frac{kq}{r_2} + \frac{k(-q)}{r_1} = kq\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$$
Since $r \gg a$, neglect $a^2\cos^2\theta$ compared to $r^2$, and $p = q\cdot 2a$:
★ Result
$$V = \frac{kp\cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{p\cos\theta}{r^2}$$
Axial Point ($\theta = 0°$)
$\cos 0° = 1$, so:
★
$$V_{axial} = \frac{kp}{r^2}$$
Maximum potential
Equatorial Point ($\theta = 90°$)
$\cos 90° = 0$, so:
★
$$V_{eq} = 0$$
Equatorial plane is at zero potential
Key: Unlike electric field, potential is a scalar — just add values algebraically. The equatorial plane of a dipole is always at zero potential.
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03
Relation Between Electric Field & Potential
E = −dV/dr · Field is negative gradient of potential
The electric field and potential are not independent — they are related. Field always points in the direction of decreasing potential. The field is the negative gradient of potential.
Fig 3.1 — E⃗ is perpendicular to equipotential surfaces and points toward decreasing V
Derivation
Let a unit positive charge move through small displacement $dr$ along the field direction.
①
Work done by electric field in displacement $dr$:
$$dW = \vec{E}\cdot d\vec{r} = E\,dr$$
②
This work equals the decrease in potential energy of unit charge, i.e., decrease in $V$:
$$dW = -dV \quad \Rightarrow \quad E\,dr = -dV$$
③
Dividing both sides by $dr$:
★ Result
$$E = -\frac{dV}{dr}$$
In 3D: $\vec{E} = -\nabla V = -\left(\dfrac{\partial V}{\partial x}\hat{i}+\dfrac{\partial V}{\partial y}\hat{j}+\dfrac{\partial V}{\partial z}\hat{k}\right)$
Negative sign: field points from high V to low V — always in the direction of decreasing potential. |
Unit: $1\;\text{V/m} = 1\;\text{N/C}$.
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04
Potential Energy of a System of Two Point Charges
U = kq₁q₂/r · Electrostatic P.E.
Potential energy of a system of charges is the work done in assembling the charges from infinity to their respective positions. It is stored in the system as electrostatic energy.
Fig 4.1 — $q_2$ brought from infinity to distance $r$ from $q_1$ (already fixed)
Derivation
Fix $q_1$ at its position (no work needed). Now bring $q_2$ from $\infty$ to a distance $r$ from $q_1$.
①
The potential at distance $r$ due to $q_1$ alone:
$$V_1(r) = \frac{kq_1}{r}$$
②
Work done to bring $q_2$ from $\infty$ to this point against the field of $q_1$:
$$W = q_2 \times V_1(r) = q_2 \times \frac{kq_1}{r}$$
③
This work is stored as the potential energy of the system:
★ Result
$$U = \frac{kq_1 q_2}{r} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r}$$
Like charges ($q_1 q_2 > 0$): $U > 0$ → repulsive, system has positive energy. |
Unlike charges ($q_1 q_2 < 0$): $U < 0$ → attractive, system is bound.
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05
Combination of Capacitors
Parallel · Series · Equivalent Capacitance
Capacitors can be combined in parallel (same voltage, charges add) or in series (same charge, voltages add). The equivalent capacitance replaces the combination.
Case I — Capacitors in Parallel
Fig 5.1 — Parallel combination: same voltage V across each capacitor
Derivation — Parallel
All capacitors share the same potential difference $V$. Charges are different.
Total voltage: $V = V_1+V_2+V_3$
$$V = Q\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)$$
③
For equivalent capacitor $C_S$: $V = \dfrac{Q}{C_S}$, so:
★ Result
$$\frac{1}{C_S} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \quad \text{(in general: }\frac{1}{C_S}=\sum\frac{1}{C_i}\text{)}$$
Equivalent capacitance is always less than the smallest individual one
Memory trick: Parallel → capacitances add (like resistors in series). Series → reciprocals add (like resistors in parallel). Opposite to resistors!
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06
Energy Stored in a Capacitor
U = ½CV² · Work done during charging
When a capacitor is charged, work is done by the battery. This work is stored as electrostatic potential energy in the electric field between the plates.
Fig 6.1 — Capacitor plates, field E⃗ between them, voltage V; energy is stored in the field
Derivation
Suppose at any instant, charge on capacitor is $q$ and voltage is $v = q/C$.
①
Small work done to add a tiny charge $dq$ against the existing voltage $v$:
$$dW = v\,dq = \frac{q}{C}\,dq$$
②
Total work done to charge from $0$ to $Q$:
$$W = \int_0^Q \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{q^2}{2}\Bigg|_0^Q = \frac{Q^2}{2C}$$
③
This work is stored as potential energy $U$. Using $Q = CV$:
★ Result
$$U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$$
Energy Density
Energy stored per unit volume between plates:
$$u = \frac{1}{2}\varepsilon_0 E^2$$
Energy lives in the electric field itself
Which formula to use?
Use $\dfrac{1}{2}CV^2$ when $C$ & $V$ are known.
Use $\dfrac{Q^2}{2C}$ when $C$ & $Q$ are known.
Use $\dfrac{1}{2}QV$ when both $Q$ & $V$ are given.
On disconnecting battery: $Q$ stays constant → $V$ & $U$ change when $C$ changes.
On keeping battery: $V$ stays constant → $Q$ & $U$ change when $C$ changes.
