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
1
$t = 0$: $V_C = 0$, $I = V_0/R$ (maximum current, capacitor uncharged)
2
$t = 1\tau$: $V_C = 63.2\%$ of $V_0$, $I = 36.8\%$ of $I_0$
3
$t = 2\tau$: $V_C = 86.5\%$ of $V_0$, $I = 13.5\%$ of $I_0$
4
$t = 3\tau$: $V_C = 95.0\%$ of $V_0$, $I = 5.0\%$ of $I_0$
5
$t = 5\tau$: $V_C \approx V_0$ (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%)
Application Strategy
Worked Examples
Problem Statement
Two opposite point charges, each of magnitude $5 \times 10^{-6}$ C, are separated by a distance of 20 cm. What is the electric potential at the midpoint between them?
Point Charge PotentialSuperposition Principle
Strategic Toolkit
$$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$
Step 1
Identify the geometry
The midpoint is at distance d = 10 cm = 0.1 m from each charge. One charge is positive (+q) and the other is negative (−q).
$$r = \frac{20}{2} = 10 \text{ cm} = 0.1 \text{ m}$$
Step 2
Calculate individual potentials
Compute the potential due to each charge at the midpoint using V = kq/r. Since the charges are equal and opposite, their potentials are equal in magnitude but opposite in sign.
$$V^+ = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 4.5 \times 10^5 \text{ V}$$
Step 3
Apply superposition
The total potential is the algebraic sum: V = V⁺ + V⁻. Since V⁻ = −V⁺, the two cancel out exactly.
$$V = 4.5 \times 10^5 + (-4.5 \times 10^5) = 0 \text{ V}$$
Problem Statement
A charge of $3 \times 10^{-8}$ C is placed in vacuum. Calculate the electric potential at a point 30 cm from the charge.
Point Charge Potential
Strategic Toolkit
$$V_r = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$
Step 1
Identify given values
Source charge q = 3 × 10⁻⁸ C, distance r = 30 cm = 0.3 m. Coulomb's constant k = 9 × 10⁹ N·m²/C².
$$q = 3 \times 10^{-8} \text{ C}, \quad r = 0.3 \text{ m}$$
Step 2
Substitute into the formula
Apply V = kq/r using the identified values.
$$V = \frac{9 \times 10^9 \times 3 \times 10^{-8}}{0.3}$$
Step 3
Calculate the result
Numerator: 9 × 10⁹ × 3 × 10⁻⁸ = 27 × 10¹ = 270. Dividing by 0.3 gives 900 volts.
$$V = \frac{27}{0.3} = 900 \text{ V}$$
Problem Statement
A $100 \text{ \mu F}$ capacitor is charged through a $50 \text{ k}\Omega$ resistor by a 12 V battery. Find the voltage across the capacitor after 5 seconds.
Capacitor ChargingTime Constant
Strategic Toolkit
$$V_C = V_0 \left(1 - e^{-t/RC}\right)$$
Step 1
Calculate the time constant
Convert to SI: R = 50 × 10³ Ω, C = 100 × 10⁻⁶ F. Multiply to find τ.
$$\tau = RC = (50 \times 10^3)(100 \times 10^{-6}) = 5 \text{ s}$$
Step 2
Evaluate the exponential
At t = 5 s, the ratio t/τ = 1, so e raised to −1 equals approximately 0.368.
$$e^{-t/RC} = e^{-5/5} = e^{-1} \approx 0.368$$
Step 3
Calculate capacitor voltage
Substitute into V_C = V₀(1 − e⁻έ/ὣ) with V₀ = 12 V.
$$V_C = 12(1 - 0.368) = 12 \times 0.632 = 7.58 \text{ V}$$
Problem Statement
A $50 \text{ \mu F}$ capacitor charged to 200 V discharges through a $100 \text{ k}\Omega$ resistor. Find the current flowing through the resistor at $t = 3$ s.
Capacitor DischargingTime Constant
Strategic Toolkit
$$I = \frac{V_0}{R} \, e^{-t/RC}$$
Step 1
Calculate the time constant
R = 100 kΩ = 10⁵ Ω, C = 50 μF = 50 × 10⁻⁶ F.
$$\tau = RC = (10^5)(50 \times 10^{-6}) = 5 \text{ s}$$
Step 2
Find the initial current
At t = 0 the full 200 V appears across the resistor, giving the initial current I₀ = V₀/R.
$$I_0 = \frac{200}{10^5} = 2 \times 10^{-3} \text{ A} = 2 \text{ mA}$$
Step 3
Evaluate the exponential at t = 3 s
The ratio t/τ = 3/5 = 0.6, and e raised to −0.6 is approximately 0.549.
$$e^{-3/5} = e^{-0.6} \approx 0.549$$
Step 4
Calculate current at t = 3 s
Multiply the initial current by the exponential decay factor.
$$I = 2 \times 0.549 = 1.098 \text{ mA}$$
Problem Statement
A $200 \text{ \mu F}$ capacitor charges through a $20 \text{ k}\Omega$ resistor from a 10 V supply. How long does it take for the capacitor voltage to reach 8 V?
Capacitor ChargingTime Constant
Strategic Toolkit
$$V_C = V_0 \left(1 - e^{-t/RC}\right)$$
Step 1
Calculate the time constant
R = 20 kΩ = 2 × 10⁴ Ω, C = 200 μF = 2 × 10⁻⁴ F.
$$\tau = RC = (2 \times 10^4)(2 \times 10^{-4}) = 4 \text{ s}$$
Step 2
Rearrange the charging formula for time
From V_C = V₀(1 − e⁻έ/τ), solve for t. Dividing both sides by V₀ and rearranging: e⁻έ/τ = 1 − V_C/V₀, which gives t = −τ ln(1 − V_C/V₀).
$$t = -\tau \ln\left(1 - \frac{V_C}{V_0}\right)$$
Step 3
Substitute values and compute
V_C = 8 V, V₀ = 10 V, so 1 − V_C/V₀ = 1 − 0.8 = 0.2. The natural log of 0.2 is approximately −1.609.
$$t = -4 \times \ln(0.2) = -4 \times (-1.609) = 6.44 \text{ s}$$
Memorize formulas
Electric Potential
Potential at a point is work done per unit charge from infinity to that point. Unit: volt (V) = 1 J/C.
Formula
$$V = \frac{W}{q_0}$$
Potential Due to Point Charge
Potential at distance r from point charge q. Use k = 1/(4πε₀) ≈ 9 × 10⁹ N·m²/C². Positive q gives positive potential; negative q gives negative potential.
Formula
$$V_r = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$
Electric Field as Potential Gradient
Electric field equals negative potential gradient. Units: V/m = N/C. Field points from high to low potential.
Formula
$$E = -\frac{\Delta V}{\Delta r}$$
RC Time Constant
Time constant τ = RC determines charging/discharging speed. At t = τ: charging reaches 63.2%, discharging drops to 36.8%. Practically complete at 5τ.
Formula
$$\tau = RC$$
Capacitor Charging Voltage
Voltage across charging capacitor rises from 0 toward V₀. At t = τ: V_C ≈ 63.2% of V₀.
Formula
$$V_C = V_0 \left(1 - e^{-t/RC}\right)$$
Capacitor Discharging Voltage
Voltage across discharging capacitor decays from V₀ toward 0. At t = τ: V_C ≈ 36.8% of V₀.
Formula
$$V_C = V_0 \, e^{-t/RC}$$
Review
1
Which of the following correctly defines electric potential at a point in an electric field?
Ans
The work done per unit charge in bringing a small positive test charge from infinity to that point
Review
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Definition of Electric Potential
Electric potential at a point equals the work done by an external agent per unit charge to bring a small positive test charge from infinity to that point.
$$V = \frac{W}{q_0}$$
Unit Analysis
Work has the unit joule (J) and charge has the unit coulomb (C), so potential has the unit J/C, which is the volt (V).
$$1 \text{ V} = 1 \text{ J/C}$$
2
Which of the following correctly expresses the volt in terms of base SI units?
Ans
J \u00b7 C\u207b\u00b9
Review
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Relate Volt to Work and Charge
From the definition V = W/q, the volt is the unit of work per unit of charge.
$$1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}$$
Verify Dimensions
Work (J) = N \u00b7 m = kg \u00b7 m\u00b2/s\u00b2, Charge (C) = A \u00b7 s, so V = kg \u00b7 m\u00b2 \u00b7 s\u207b\u00b3 \u00b7 A\u207b\u00b1 in base units.
$$\text{V} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}$$
3
The relationship between electric field and potential is given by E = −ΔV/Δr. What does the negative sign indicate?
Ans
The electric field points in the direction of decreasing potential
Review
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Electric Field as Negative Potential Gradient
The electric field at a point equals the negative of the rate of change of potential with respect to displacement in that direction.
$$E = -\frac{\Delta V}{\Delta r}$$
Interpret the Negative Sign
The negative sign means the electric field vector is directed from higher potential toward lower potential. It does not constrain the potential to always decrease with distance.
$$\vec{E} \text{ points along } -\nabla V$$
4
A point charge of +4 \u03bcC is placed in free space. What is the electric potential at a distance of 30 cm from the charge? (k = 9 \u00d7 10\u2079 N \u00b7 m\u00b2/C\u00b2)
Ans
1.2 \u00d7 10\u2075 V
Review
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Identify Given Values
q = 4 \u00d7 10\u207b\u2076 C, r = 30 cm = 0.3 m, k = 9 \u00d7 10\u2079 N \u00b7 m\u00b2/C\u00b2
$$q = 4 \times 10^{-6} \text{ C}, \quad r = 0.3 \text{ m}$$
Apply Point Charge Potential Formula
Substitute values into V = kq/r.
$$V = \frac{(9 \times 10^9)(4 \times 10^{-6})}{0.3}$$
Calculate Result
V = 36 \u00d7 10\u00b3 / 0.3 = 1.2 \u00d7 10\u2075 V
$$V = 1.2 \times 10^5 \text{ V}$$
5
Two point charges, +6 \u03bcC and −6 \u03bcC, are placed 40 cm apart. What is the electric potential at the midpoint between them? (k = 9 \u00d7 10\u2079 N \u00b7 m\u00b2/C\u00b2)
Ans
0 V
Review
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Determine Distance to Each Charge
The midpoint is equidistant from both charges: r = 40/2 = 20 cm = 0.2 m.
$$r = 0.2 \text{ m}$$
Calculate Potential from Each Charge
V\u208a = kq\u208a/r and V\u208b = kq\u208b/r. Since potential is scalar, we add them algebraically.
$$V_+ = \frac{(9 \times 10^9)(6 \times 10^{-6})}{0.2} = 2.7 \times 10^5 \text{ V}$$
Sum the Potentials
V\u208b = −2.7 \u00d7 10\u2075 V, so V\u209c\u2092\u209c = 2.7 \u00d7 10\u2075 + (−2.7 \u00d7 10\u2075) = 0 V.
$$V_{\text{total}} = 0 \text{ V}$$
6
A 100 \u03bcF capacitor is connected in series with a 50 k\u03a9 resistor. What is the time constant of the RC circuit?
Ans
5.0 s
Review
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Write the Time Constant Formula
The time constant for an RC circuit equals the product of resistance and capacitance.
$$\tau = RC$$
Substitute and Calculate
R = 50 \u00d7 10\u00b3 \u03a9, C = 100 \u00d7 10\u207b\u2076 F
$$\tau = (50 \times 10^3)(100 \times 10^{-6}) = 5.0 \text{ s}$$
7
A 200 \u03bcF capacitor is connected to a 12 V battery through a 100 k\u03a9 resistor. What is the voltage across the capacitor after one time constant?
Ans
7.6 V
Review
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Apply Charging Voltage Formula at t = \u03c4
After one time constant, t/RC = 1, so the exponential term becomes e\u207b\u00b9 \u2248 0.368.
$$V_C = V_0 \left(1 - e^{-t/RC}\right) = 12(1 - e^{-1})$$
Compute the Voltage
e\u207b\u00b9 \u2248 0.368, so V_C = 12 \u00d7 (1 − 0.368) = 12 \u00d7 0.632
$$V_C \approx 7.6 \text{ V}$$
8
A capacitor charged to 20 V is allowed to discharge through a 10 k\u03a9 resistor. After two time constants, what is the voltage across the capacitor?
Ans
2.7 V
Review
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Apply Discharging Formula
After t = 2\u03c4, the exponential term is e\u207b\u00b2.
$$V_C = V_0 \, e^{-t/RC} = 20 \times e^{-2}$$
Evaluate Exponential
e\u207b\u00b2 \u2248 0.1353
$$V_C = 20 \times 0.1353 \approx 2.7 \text{ V}$$
9
A capacitor is being charged through a resistance by a DC source. After a very long time (many time constants), what will be the voltage across the capacitor?
Ans
Equal to the source voltage
Review
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Take the Limit of Charging Formula
As t approaches infinity, the exponential term e\u207b\u02d2/\u1d63\u1d04 approaches zero.
$$\lim_{t \to \infty} V_C = V_0(1 - 0) = V_0$$
Steady State Condition
In steady state, the capacitor is fully charged, current drops to zero, and the full source voltage appears across the capacitor.
$$V_C = V_0 \text{ (steady state)}$$
10
A 50 \u03bcF capacitor is charged through a 200 k\u03a9 resistor by a 10 V battery. How long does it take for the capacitor voltage to reach 6.32 V?
Ans
10.0 s
Review
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Set Up the Charging Equation
6.32 = 10(1 − e\u207b\u02d2/\u1d63\u1d04)
$$0.632 = 1 - e^{-t/RC}$$
Solve for the Exponential
e\u207b\u02d2/\u1d63\u1d04 = 1 − 0.632 = 0.368 = e\u207b\u00b9, so t/RC = 1.
$$\frac{t}{RC} = 1 \implies t = RC$$
Calculate the Time Constant
R = 200 k\u03a9 = 2 \u00d7 10\u2075 \u03a9, C = 50 \u03bcF = 5 \u00d7 10\u207b\u2075 F
$$t = (2 \times 10^5)(5 \times 10^{-5}) = 10 \text{ s}$$
11
The electric potential at a distance r from an isolated point charge is V. If the distance from the charge is doubled to 2r, what happens to the potential?
Ans
It becomes V/2
Review
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State the Original Potential
At distance r, the potential is V = kq/r.
$$V = \frac{kq}{r}$$
Apply the Doubled Distance
At distance 2r, V' = kq/(2r) = (1/2)(kq/r) = V/2.
$$V' = \frac{kq}{2r} = \frac{V}{2}$$
12
During the charging of a capacitor through a resistor from a DC supply, how does the charging current behave over time?
Ans
It decreases exponentially from an initial maximum toward zero
Review
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Write the Charging Current Expression
The current through the resistor during charging is proportional to the remaining voltage difference divided by R.
$$I(t) = \frac{V_0}{R} \, e^{-t/RC}$$
Analyze Current Behavior
At t = 0: I = V\u2080/R (maximum). As t increases, e\u207b\u02d2/\u1d63\u1d04 decays toward zero, so I decays exponentially toward zero.
$$\text{Current decays exponentially from } I_{\max} \text{ to } 0$$
13
A 200 \u03bcF capacitor charged to 10 V discharges through a 25 k\u03a9 resistor. After approximately how many seconds will the voltage across the capacitor drop to 1.35 V?
Ans
10.0 s
Review
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Set Up Discharge Equation
1.35 = 10 \u00d7 e\u207b\u02d2/\u1d63\u1d04
$$e^{-t/RC} = 0.135$$
Solve for Time
ln(0.135) \u2248 −2, so t/RC = 2, meaning t = 2\u03c4.
$$t = 2RC$$
Calculate Using Given Values
RC = (25 \u00d7 10\u00b3)(200 \u00d7 10\u207b\u2076) = 5 s, so t = 2 \u00d7 5 = 10 s.
$$t = 10 \text{ s}$$
14
At a certain point in a region of space, the electric potential increases uniformly in the +x direction. What is the direction of the electric field at that point?
Ans
−x direction
Review
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Apply the Potential Gradient Relationship
The electric field is the negative gradient of potential. If V increases in +x, then dV/dx > 0.
$$E_x = -\frac{\Delta V}{\Delta x}$$
Determine Direction
Since \u0394V/\u0394x > 0, E_x is negative, meaning E points in the −x direction.
$$\vec{E} \text{ points in the } -x \text{ direction}$$
15
Which of the following modifications to an RC circuit will result in doubling the time constant while keeping all other parameters the same?
Ans
Doubling the resistance only
Review
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State the Time Constant
The time constant depends on resistance and capacitance only.
$$\tau = RC$$
Evaluate Each Option
Double R: \u03c4' = 2RC (doubles). Halve C: \u03c4' = RC/2 (halves). Double V: \u03c4' = RC (no change). Halve both: \u03c4' = RC/4 (quarters).
$$\text{Doubling } R \text{ gives } \tau' = 2\tau$$
Question 1 of 15
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1
Which of the following correctly defines electric potential at a point in an electric field?