Electric Potential & Capacitor Transients
Electric Potential and Potential Difference
When a positive charge $q_0$ is placed in an Electric Field, it experiences a force and gains kinetic energy if allowed to move along the field direction. To move the charge against the field, an external force equal and opposite to $q_0 \mathbf{E}$ must be applied. The work done by this external force increases the electrical potential energy of the charge. The Electric Potential at any point is defined as the work done per unit charge in bringing a small positive test charge from infinity to that point while maintaining electrostatic equilibrium (uniform velocity).
$$V = \frac{W}{q_0}$$
Electric potential at a point equals the work done per unit charge in bringing a positive test charge from infinity to that point
$V$=Electric potential at the point(V (volt))
$W$=Work done in moving the test charge from infinity(J (joule))
$q_0$=Magnitude of the test charge(C (coulomb))
$q_0 = 1$ C
→Potential numerically equals the work done in joules
Reference at Infinity: By convention, electric potential is taken as zero at infinity. Potential at any point is thus a potential difference between that point and infinity.
Scalar Quantity: Both $W$ and $q_0$ are scalars, so electric potential is a scalar quantity. It can be positive, negative, or zero depending on the source charge.
Work-Potential Relation: The change in potential energy of a charge $q_0$ moved between two points A and B is $\Delta U = q_0 \Delta V = W_{AB}$
The Potential Difference between two points A and B is defined as the work done per unit charge in moving a positive charge from A to B while maintaining equilibrium. It is one of the most fundamental quantities in electrostatics and circuit analysis, directly determining the energy gained or lost by charges moving between the two points.
$$\Delta V = V_B - V_A = \frac{W_{AB}}{q_0}$$
Potential difference between two points equals the work done per unit charge between those points
$V_B - V_A$=Potential difference from A to B(V (volt))
$W_{AB}$=Work done by external force moving charge from A to B(J (joule))
$q_0$=Test charge magnitude(C (coulomb))
$V_B > V_A$
→Work is done by the external force against the field
$V_B < V_A$
→The field does work on the charge as it moves from A to B
Volt Definition: The Volt is the SI unit of potential difference — one volt exists between two points when one joule of work is done in moving one coulomb of charge between them: $1 \text{ V} = 1 \text{ J/C}$
Relation to Energy: The potential energy change is directly related to potential difference: $\Delta U = q_0 \Delta V$. A charge gains energy moving to higher potential and loses energy moving to lower potential.
Practical Significance: Potential difference drives current in circuits. A 12 V battery does 12 J of work per coulomb of charge moved through the circuit.
Electric Field as Potential Gradient
For a uniform electric field between two parallel plates separated by distance $d$, moving a charge against the field requires work $W_{AB} = Fd = -q_0 Ed$, where the negative sign arises because the external force must oppose $q_0 E$ to maintain equilibrium. Substituting into the Potential Difference definition gives the fundamental relation between Electric Field intensity and Potential Gradient.
$$E = -\frac{\Delta V}{d} = -\frac{\Delta V}{\Delta r}$$
Electric field equals the negative of the potential gradient — the rate of change of potential with distance
$E$=Electric field intensity(V/m or N/C)
$\Delta V$=Change in electric potential(V (volt))
$d$ or $\Delta r$=Displacement along the field direction(m (metre))
$\Delta V > 0$ (potential increases)
→The field direction is opposite to the direction of increasing potential
Field lines are closer together
→Potential changes more rapidly per unit distance, indicating a stronger field
Negative Sign: The negative sign indicates that the Electric Field always points from higher to lower potential. Field lines are directed along the direction of decreasing potential.
Potential Gradient: The quantity $\Delta V / \Delta r$ is the Potential Gradient. Its maximum value occurs along a field line where the distance between equipotential surfaces is minimum.
Unit Equivalence: The unit V/m equals N/C: $1 \text{ V/m} = 1 \text{ (J/C)/m} = 1 \text{ (N \cdot m/C)/m} = 1 \text{ N/C}$
Potential Due to a Point Charge
To derive the Electric Potential at distance $r$ from a Point Charge $q$, we bring a unit positive test charge from infinity to that point while maintaining equilibrium. Since the field of a point charge varies as $1/r^2$, it is not constant along the path, so we use calculus. Consider two closely spaced points A and B at distances $r_A$ and $r_B$ from $q$, with midpoint distance $r$. For infinitesimally close points, $r^2 \approx r_A r_B$. The work done in moving a unit charge from B to A gives the potential difference, and letting B approach infinity yields the absolute potential.
$$V_r = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$
Electric potential at distance r from a point charge q; potential decreases with distance and its sign depends on the source charge
$V_r$=Electric potential at distance r(V (volt))
$q$=Source point charge(C (coulomb))
$r$=Distance from the point charge(m (metre))
$\varepsilon_0$=[Permittivity of Free Space], $8.85 \times 10^{-12}$ C\u00b2/N\u00b7m\u00b2(C\u00b2/N\u00b7m\u00b2)
$q > 0$
→Potential is positive — work is done against the repulsive field to bring a positive test charge from infinity
$q < 0$
→Potential is negative — the attractive field does work on the test charge
$r \to \infty$
→Potential approaches zero, consistent with the reference convention
Derivation Summary: For two closely spaced points at $r_A$ and $r_B$: $V_A - V_B = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{r_A} - \frac{1}{r_B} \right)$. Setting $r_B \to \infty$ (so $V_B = 0$) gives $V_A = \frac{q}{4\pi\varepsilon_0 r_A}$.
Inverse Distance Law: Unlike electric field ($\propto 1/r^2$), potential varies as $1/r$. This is because potential is the integral (work) of the field over distance.
Superposition Principle: The total potential at a point due to multiple charges is the algebraic sum of individual potentials: $V = V_1 + V_2 + V_3 + \cdots$. Since potential is scalar, signs are added directly.
Midpoint Between Opposite Charges: At the midpoint between equal and opposite charges $+q$ and $-q$ separated by distance $2d$, the net potential is zero: $V = \frac{q}{4\pi\varepsilon_0 d} + \frac{-q}{4\pi\varepsilon_0 d} = 0$
Electric Field vs Electric Potential (Point Charge)
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Electric field: $E = \frac{q}{4\pi\varepsilon_0 r^2}$ — inversely proportional to $r^2$
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Electric potential: $V = \frac{q}{4\pi\varepsilon_0 r}$ — inversely proportional to $r$
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Electric field is a vector quantity; potential is a scalar quantity
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Electric field can be zero where potential is non-zero (e.g., midpoint between like charges)
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Potential can be zero where field is non-zero (e.g., midpoint between opposite charges)
Capacitor Charging Through Resistance
When a Capacitor is connected in series with a resistor and a battery of emf $V_0$, the capacitor does not charge instantaneously. As charge builds up on the plates, the growing voltage across the capacitor opposes the battery, reducing the current. This produces a Transient response where the charging current starts at a maximum and decays exponentially until the capacitor reaches full charge and the current ceases.
$$V_C = V_0 \left( 1 - e^{-t/RC} \right)$$
Voltage across a charging capacitor increases exponentially from zero toward the supply voltage V₀
$V_C$=Voltage across the capacitor at time t(V (volt))
$V_0$=Supply voltage (battery emf)(V (volt))
$R$=Series resistance(Ω (ohm))
$C$=Capacitance(F (farad))
$t$=Time since the switch was closed(s (second))
$t = 0$
→$V_C = 0$ — capacitor uncharged, all voltage appears across the resistor
$t \to \infty$
→$V_C = V_0$ — capacitor fully charged, current is zero
$t = RC$
→$V_C \approx 0.632 V_0$ — capacitor has reached 63.2% of its final voltage
Charging Current: The current during charging decays exponentially: $I = \frac{V_0}{R} e^{-t/RC}$. At $t = 0$, the current is maximum ($I_0 = V_0/R$) because the uncharged capacitor initially acts as a short circuit.
Charge Build-up: The charge on the plates grows as $q = q_0(1 - e^{-t/RC})$ where $q_0 = CV_0$ is the maximum charge on the fully charged capacitor.
The product $RC$ is called the Time Constant ($\tau$) of the circuit. It is the fundamental parameter that determines how quickly a Capacitor charges or discharges. A larger resistance or larger capacitance means the process takes longer, as more time is needed to build up or drain the required charge through the circuit.
$$\tau = RC$$
Time constant of an RC circuit — the time for a charging capacitor to reach 63.2% of its final voltage, or for a discharging capacitor to fall to 36.8% of its initial voltage
$\tau$=Time constant(s (second))
$R$=Resistance(Ω (ohm))
$C$=Capacitance(F (farad))
$t = \tau$
→Charging: 63.2% complete; Discharging: 36.8% of initial voltage remains
$t = 5\tau$
→Process is 99.3% complete — effectively at Steady State
$R$ is large
→Current is limited, so charging and discharging proceed more slowly
Physical Meaning: After one Time Constant, the capacitor has acquired 63.2% of its maximum charge during charging (or lost 63.2% during discharging). After $5\tau$, the process is effectively complete at 99.3%.
Dimensional Check: $RC$ has units of $\Omega \times \text{F} = (\text{V/A}) \times (\text{C/V}) = \text{C/A} = \text{s}$, confirming it has dimensions of time.
Practical Example: For $R = 10 \text{ k}\Omega$ and $C = 100 \text{ \mu F}$, $\tau = 10^4 \times 10^{-4} = 1$ s. Full charge takes approximately 5 seconds.
Charging Progress at Key Time Intervals
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$t = 0$: $V_C = 0$, $I = V_0/R$ (maximum current, capacitor uncharged)
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$t = 1\tau$: $V_C = 63.2\%$ of $V_0$, $I = 36.8\%$ of $I_0$
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$t = 2\tau$: $V_C = 86.5\%$ of $V_0$, $I = 13.5\%$ of $I_0$
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$t = 3\tau$: $V_C = 95.0\%$ of $V_0$, $I = 5.0\%$ of $I_0$
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$t = 5\tau$: $V_C \approx V_0$ (fully charged), $I \approx 0$
Capacitor Discharging Through Resistance
When a charged Capacitor with initial voltage $V_0$ is disconnected from the supply and discharges through a resistor $R$, the stored energy drives current through the circuit. The voltage across the capacitor, the charge on its plates, and the current all decay exponentially from their initial values toward zero. Like charging, this is a Transient process that reaches Steady State after approximately $5\tau$ when the capacitor is fully discharged and the current ceases.
$$V_C = V_0 \, e^{-t/RC}$$
Voltage across a discharging capacitor decays exponentially from its initial value V₀ toward zero
$V_C$=Voltage across capacitor at time t(V (volt))
$V_0$=Initial voltage across capacitor at t = 0(V (volt))
$RC$=Time constant $\tau = RC$(s (second))
$t = 0$
→$V_C = V_0$ — capacitor fully charged
$t = RC$
→$V_C \approx 0.368 V_0$ — 36.8% of initial voltage remains
$t \to \infty$
→$V_C = 0$ — capacitor fully discharged
Discharge Current: The discharge current is $I = -\frac{V_0}{R} e^{-t/RC}$. The negative sign indicates the current flows in the opposite direction to the charging current. Its magnitude decays from $V_0/R$ to zero exponentially.
Charging vs Discharging Summary
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Charging: $V_C = V_0(1 - e^{-t/RC})$ — voltage rises from 0 to $V_0$
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Discharging: $V_C = V_0 \, e^{-t/RC}$ — voltage falls from $V_0$ to 0
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Charging current: $I = \frac{V_0}{R} e^{-t/RC}$ — decays from $I_0$ to 0
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Discharge current: $I = -\frac{V_0}{R} e^{-t/RC}$ — decays from $-I_0$ to 0 (opposite direction)
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Both processes share the same Time Constant $\tau = RC$
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Both processes are effectively complete after $t = 5\tau$ (99.3%)