Electric potential

1. Overview

Electric potential is a fundamental scalar quantity that describes the energy environment of an electric field. While electric field strength ($\mathbf{E}$) is a vector quantity describing the force per unit charge, electric potential ($V$) describes the work done per unit charge. This shift from force to energy allows for the application of the Principle of Conservation of Energy to electrostatic systems. By understanding the "potential landscape" created by charges, we can predict the motion and final velocities of charged particles, such as electrons in a CRT or alpha particles in scattering experiments, without needing to track complex force vectors at every point in space.

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Key Definitions

• Electric Potential ($V$): The work done per unit positive charge in bringing a small test charge from infinity to a point.
• Electric Potential Energy ($E_P$): The work done in bringing a charge from infinity to a point in an electric field.
• Potential Gradient: The change in electric potential per unit change in distance at a point in an electric field.
• Equipotential Surface: A surface where all points are at the same electric potential. No work is done when a charge moves along such a surface, and electric field lines always meet these surfaces at right angles ($90^\circ$).

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Content

3.1 Defining Electric Potential ($V$)

Electric potential is a scalar quantity. This is a significant advantage in physics problems because, unlike electric field strength, potentials from multiple charges can be added using simple algebraic addition rather than vector resolution.

The unit of electric potential is the Volt (V), which is defined as one Joule per Coulomb ($J\,C^{-1}$).

The Significance of Infinity: In electrostatics, we define the zero-point of potential ($V = 0$) at infinity. This is because as the distance $r$ from a source charge becomes infinitely large, the force exerted by that charge becomes zero. By using infinity as a universal reference point, we can determine the absolute potential at any specific point $r$.

• Positive Potential: Exists around an isolated positive charge. Work must be done against the electrostatic repulsion to bring a positive test charge from infinity.
• Negative Potential: Exists around an isolated negative charge. The field does work on the positive test charge as it is attracted from infinity; therefore, the potential is lower than the potential at infinity (which is zero).

3.2 Potential Due to a Point Charge

For an isolated point charge $Q$, the electric potential $V$ at a distance $r$ from the charge is derived from the inverse square law of the electric field. The formula is:

$$V = \frac{Q}{4\pi\varepsilon_0r}$$ (This equation is provided on the Formulae Sheet)

Where:

• $V$ = Electric potential ($V$ or $J\,C^{-1}$)
• $Q$ = The charge creating the field ($C$)
• $\varepsilon_0$ = Permittivity of free space ($\approx 8.85 \times 10^{-12}\,F\,m^{-1}$)
• $r$ = Distance from the center of the charge ($m$)

Graphical Representation:

1. $V$ against $r$: The graph follows a $1/r$ relationship. For a positive charge, the curve is in the first quadrant, decreasing towards the r-axis. For a negative charge, the curve is in the fourth quadrant, increasing from a large negative value towards zero.
2. Comparison with $E$: While $E \propto 1/r^2$ (inverse square), $V \propto 1/r$ (inverse). This means the potential decreases more slowly with distance than the field strength does.

3.3 Potential of a Charged Isolated Sphere

A fundamental principle in the Cambridge 9702 syllabus is the behavior of a charged conducting sphere.

• Outside the sphere ($r > R$): The sphere behaves as a point charge where all charge $Q$ is concentrated at the center. The formula $V = \frac{Q}{4\pi\varepsilon_0r}$ applies.
• On the surface ($r = R$): The potential is $V = \frac{Q}{4\pi\varepsilon_0R}$.
• Inside the sphere ($r < R$): The electric field strength $E$ inside a conductor is zero. Since $E = -dV/dr$, if $E = 0$, then the gradient of the potential is zero. This means the potential is constant everywhere inside the sphere and is equal to the potential at the surface.

3.4 The Relationship Between $E$ and $V$

The electric field strength at a point is numerically equal to the negative of the potential gradient at that point.

$$E = -\frac{dV}{dr}$$ (This equation must be memorised)

Key Interpretations:

1. The Gradient: On a graph of $V$ against $r$, the gradient at any point is equal to $-E$. To find $E$ at a specific distance, draw a tangent to the $V-r$ curve and calculate its slope.
2. The Negative Sign: This indicates that the electric field $\mathbf{E}$ points in the direction of decreasing potential. Just as a ball rolls down a gravitational hill (high to low potential), a positive charge is pushed "down" the electric potential gradient.
3. Uniform Fields: In a uniform field (e.g., between parallel plates), the gradient is constant. Therefore: $$E = \frac{\Delta V}{\Delta d}$$ (Where $\Delta V$ is the potential difference and $\Delta d$ is the separation).

3.5 Electric Potential Energy ($E_P$)

Electric potential energy is the energy stored in a system of charges due to their relative positions. If a charge $q$ is placed at a point where the electric potential is $V$, its potential energy is:

$$E_P = qV$$

For a system consisting of two point charges $Q$ and $q$ separated by a distance $r$, the electric potential energy is:

$$E_P = \frac{Qq}{4\pi\varepsilon_0r}$$ (This equation is provided on the Formulae Sheet)

Energy and Signs:

• Like charges ($+Q, +q$ or $-Q, -q$): $E_P$ is positive. Work must be done to push them together. As $r$ decreases, $E_P$ increases.
• Opposite charges ($+Q, -q$): $E_P$ is negative. The charges are bound together. Work must be done to pull them apart (to move them to infinity where $E_P = 0$). As $r$ decreases, $E_P$ becomes more negative (decreases).

3.6 Conservation of Energy in Electric Fields

When a charged particle moves freely in an electric field, its total energy remains constant. The change in electric potential energy is converted into a change in kinetic energy.

$$\Delta E_P + \Delta E_K = 0$$ $$q\Delta V = \frac{1}{2}m(v^2 - u^2)$$

This is frequently used to calculate the "speed of approach" or the "closest distance of approach" in atomic physics.

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4. Worked Examples

Worked example 1 — Closest Distance of Approach

An alpha particle ($q = +2e$, $m = 6.64 \times 10^{-27}\,kg$) is fired directly at a stationary gold nucleus ($Q = +79e$) with an initial kinetic energy of $5.0\,MeV$. Calculate the closest distance of approach.

Step 1: Convert energy to Joules. $$1.0\,eV = 1.60 \times 10^{-19}\,J$$ $$E_K = (5.0 \times 10^6) \times (1.60 \times 10^{-19}) = 8.0 \times 10^{-13}\,J$$

Step 2: Apply conservation of energy. At the closest distance ($r$), the alpha particle momentarily stops. All initial $E_K$ has been converted to $E_P$. $$E_K = \frac{Qq}{4\pi\varepsilon_0r}$$

Step 3: Rearrange for $r$ and substitute values. $$r = \frac{Qq}{4\pi\varepsilon_0 E_K}$$ $$Q = 79 \times (1.60 \times 10^{-19}) = 1.264 \times 10^{-17}\,C$$ $$q = 2 \times (1.60 \times 10^{-19}) = 3.20 \times 10^{-19}\,C$$ $$r = \frac{(1.264 \times 10^{-17}) \times (3.20 \times 10^{-19})}{4 \times \pi \times (8.85 \times 10^{-12}) \times (8.0 \times 10^{-13})}$$

Step 4: Calculate intermediate steps and final answer. $$Numerator = 4.0448 \times 10^{-36}$$ $$Denominator = 8.897 \times 10^{-23}$$ $$r = 4.546 \times 10^{-14}\,m$$ Answer: $r = 4.5 \times 10^{-14}\,m$ (to 2 s.f.)

Worked example 2 — Potential from Multiple Charges

Two point charges, $A = +4.0\,nC$ and $B = -2.0\,nC$, are placed $10\,cm$ apart in a vacuum. Calculate the electric potential at a point $P$ located on the line joining them, $4.0\,cm$ from charge $A$.

Step 1: Identify distances. Distance from $A$ to $P$ ($r_A$) = $0.04\,m$ Distance from $B$ to $P$ ($r_B$) = $0.10 - 0.04 = 0.06\,m$

Step 2: Calculate potential due to each charge. $$V_A = \frac{Q_A}{4\pi\varepsilon_0 r_A} = \frac{4.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times 0.04} = +899.2\,V$$ $$V_B = \frac{Q_B}{4\pi\varepsilon_0 r_B} = \frac{-2.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times 0.06} = -299.7\,V$$

Step 3: Sum the potentials algebraically. $$V_{total} = V_A + V_B$$ $$V_{total} = 899.2 + (-299.7) = 599.5\,V$$ Answer: $V = 600\,V$ (to 2 s.f.)

Worked example 3 — Field Strength from a $V-r$ Graph

A graph of electric potential $V$ against distance $r$ shows that at $r = 0.05\,m$, the tangent to the curve has a vertical change of $-200\,V$ over a horizontal distance of $0.02\,m$. Determine the electric field strength at this point.

Step 1: Calculate the gradient of the tangent. $$\text{Gradient} = \frac{\Delta V}{\Delta r} = \frac{-200}{0.02} = -10,000\,V\,m^{-1}$$

Step 2: Apply the relationship $E = -(\text{gradient})$. $$E = -(-10,000) = 10,000\,V\,m^{-1}$$ Answer: $E = 1.0 \times 10^4\,V\,m^{-1}$

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Key Equations

Equation	Symbols	SI Units	Source
$V = \frac{Q}{4\pi\varepsilon_0r}$	$V$: Potential, $Q$: Source charge, $r$: Distance	$V$ (V), $Q$ (C), $r$ (m)	Formula Sheet
$E = -\frac{dV}{dr}$	$E$: Field strength, $dV/dr$: Potential gradient	$E$ ($V\,m^{-1}$), $V$ (V), $r$ (m)	Memorise
$E_P = \frac{Qq}{4\pi\varepsilon_0r}$	$E_P$: Potential Energy, $Q,q$: Charges	$E_P$ (J), $Q,q$ (C), $r$ (m)	Formula Sheet
$W = q\Delta V$	$W$: Work Done, $q$: Charge, $\Delta V$: Potential Diff.	$W$ (J), $q$ (C), $V$ (V)	Memorise
$E = \frac{\Delta V}{\Delta d}$	$E$: Uniform field strength, $d$: Separation	$E$ ($V\,m^{-1}$), $V$ (V), $d$ (m)	Formula Sheet

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Common Mistakes to Avoid

• ❌ Wrong: Adding potentials using vector triangles or Pythagoras.

• ✓ Right: Potential is a scalar. Simply add the numbers (e.g., $+500V + (-200V) = +300V$).

• ❌ Wrong: Forgetting to convert distances from $cm$ or $mm$ to $m$.

• ✓ Right: Always use SI base units ($m$) in the $1/4\pi\varepsilon_0r$ formula to avoid being out by factors of 100 or 1000.

• ❌ Wrong: Assuming the potential inside a charged sphere is zero because the field is zero.

• ✓ Right: The potential is constant inside a sphere. If the surface is at $100\,V$, the center is also at $100\,V$.

• ❌ Wrong: Confusing $E \propto 1/r^2$ and $V \propto 1/r$.

• ✓ Right: Check the power of $r$ carefully. If you are calculating potential, it is never $r^2$.

• ❌ Wrong: Ignoring the sign of the charge in $E_P = qV$.

• ✓ Right: An electron ($q = -1.6 \times 10^{-19}\,C$) moving to a higher potential ($+V$) actually decreases its potential energy.

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Exam Tips

1. Definition Precision: If asked to define electric potential, you must include:

   • "Work done per unit positive charge"
   • "Bringing a small test charge"
   • "From infinity to the point" Missing "unit positive" or "infinity" are the most common reasons students lose marks.

2. Equipotential Logic: Remember that no work is done when moving a charge along an equipotential line. If an exam question shows a charge moving in a loop or a complex path but ending on the same equipotential line it started on, the net work done is zero.

3. Graph Interpretation:

   • The gradient of a $V-r$ graph is $-E$.
   • The area under an $E-r$ graph is $\Delta V$ (potential difference).

4. The "Work Done" Direction:

   • If a positive charge moves from low $V$ to high $V$, work is done on the charge (External agent).
   • If a positive charge moves from high $V$ to low $V$, work is done by the field (Kinetic energy increases).

5. Significant Figures: Cambridge 9702 usually requires answers to 2 or 3 significant figures. Since the constants like $e$ and $\varepsilon_0$ are given to 3 s.f. on the data sheet, providing 2 or 3 s.f. is standard. Never leave an answer as a fraction or with 5+ decimal places.