Electric Fields

Electric Fields

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Board Coverage AQA Paper 2 | Edexcel CP3 | OCR (A) Paper 2 | CIE P4

Charges and Fields

Explore the simulation above to develop intuition for this topic.

1. Coulomb's Law

Coulomb's Law. The electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$ in vacuum is:

$\boxed{F = \frac◆LB◆q_1 q_2◆RB◆◆LB◆4\pi\varepsilon_0 r^2◆RB◆}$

where $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$ is the permittivity of free space.

The force is attractive for opposite charges and repulsive for like charges. It acts along the line joining the charges.

Coulomb's constant: $k = \frac◆LB◆1◆RB◆◆LB◆4\pi\varepsilon_0◆RB◆ = 8.99 \times 10^9$ N m$^2$ C$^{-2}$.

Comparison with Gravity

Property	Gravitational	Electrostatic
Force law	$F = Gm_1m_2/r^2$	$F = kq_1q_2/r^2$
Constant	$G = 6.67 \times 10^{-11}$	$k = 8.99 \times 10^9$
Nature	Always attractive	Attractive or repulsive
Acts on	Mass	Charge
Relative strength	Very weak	Very strong

The electrostatic force is $\sim 10^{36}$ times stronger than gravity. We don't notice this in everyday life because most objects have nearly equal numbers of positive and negative charges.

Inverse Square Law Parallels

Both gravitational and electrostatic forces obey the inverse square law, $F \propto 1/r^2$. This shared geometry arises because both fields propagate through three-dimensional space: the surface area of a sphere grows as $4\pi r^2$, so field lines spread out and their density (which determines field strength) decreases as $1/r^2$.

The key difference is that gravitational potential is always negative (zero at infinity, becoming more negative as $r$ decreases), whereas electric potential can be positive or negative depending on the source charge. This means gravitational bound systems always have negative total energy, while electrostatic bound systems can be either bound ($U \lt 0$) or unbound ($U \gt 0$).

Concept	Gravitational Field	Electric Field
Field strength	$g = GM/r^2$ (always towards mass)	$E = kQ/r^2$ (away from $+Q$, towards $-Q$)
Potential	$V_g = -GM/r$ (always negative)	$V = kQ/r$ (sign depends on $Q$)
Potential energy	$U = -GMm/r$ (always negative)	$U = kq_1q_2/r$ (sign depends on charges)
Gradient relation	$g = -dV_g/dr$	$E = -dV/dr$
Equipotentials	Concentric spheres	Concentric spheres
Zero reference	At infinity	At infinity
Screening	None (gravity unscreened)	Conductors shield electric fields

Defining similarity. Both fields have a central $1/r$ potential, and both satisfy a Gauss's-law-type conservation of flux. The field at radius $r$ from a point source depends only on the total enclosed "charge" (mass or electric charge), not on its distribution. This means a spherically symmetric shell of charge produces the same external field as a point charge at its centre — a result exploited in both gravitational and electrostatic problems.

Difference in screening. A significant practical difference is that electric fields can be screened by conductors (Faraday cages), whereas gravitational fields cannot. This is because there are two types of electric charge that can rearrange themselves to cancel external fields, but there is only one type of mass.

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info (Paper 2). Edexcel: Explicit comparison required in CP3. OCR (A): Potential energy comparison is commonly tested (Paper 2). CIE: Gravitational--electrostatic analogies appear in P4.

2. Electric Field Strength

Definition. The electric field strength $\mathbf{E}$ at a point is the force per unit positive charge placed at that point:

$\mathbf{E} = \frac◆LB◆\mathbf{F}◆RB◆◆LB◆q◆RB◆$

SI units: N C$^{-1}$ (equivalent to V m$^{-1}$).

Field of a Point Charge

$\boxed{E = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r^2◆RB◆}$

Derivation. Place a test charge $q$ at distance $r$ from $Q$. By Coulomb's law: $F = \frac◆LB◆Qq◆RB◆◆LB◆4\pi\varepsilon_0 r^2◆RB◆$. Therefore:

$E = \frac{F}{q} = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r^2◆RB◆$

$\square$

The field points radially outward from a positive charge and radially inward towards a negative charge.

Uniform Electric Field

Between two parallel plates with p.d. $V$ and separation $d$:

$\boxed{E = \frac{V}{d}}$

Derivation. A charge $q$ moving between the plates experiences force $F = qE$. The work done is $W = Fd = qEd$. But also $W = qV$. Therefore $qEd = qV$, giving $E = V/d$. $\square$

Field of an Infinite Plane of Charge

Consider an infinite plane with uniform surface charge density $\sigma$ (C m$^{-2}$). The field on either side of the plane is:

$\boxed{E = \frac◆LB◆\sigma◆RB◆◆LB◆2\varepsilon_0◆RB◆}$

This is remarkable: the field is independent of distance from the plane. No matter how far away you move, the field strength does not decrease. This is because an infinite plane always subtends the same solid angle from any observation point.

Proof of the Infinite Plane Field

Use a cylindrical Gaussian surface of cross-sectional area $A$ piercing the plane, with equal-length extensions on each side. By symmetry, $\mathbf{E}$ is perpendicular to the plane, has equal magnitude on both sides, and has no component parallel to the plane.

Flux through both end caps: $\Phi_E = EA + EA = 2EA$. (No flux through the curved surface since $\mathbf{E}$ is parallel to it.)

Enclosed charge: $Q_{\mathrm{enc}} = \sigma A$.

By Gauss's law, $\Phi_E = Q_{\mathrm{enc}}/\varepsilon_0$:

$2EA = \frac◆LB◆\sigma A◆RB◆◆LB◆\varepsilon_0◆RB◆$

$E = \frac◆LB◆\sigma◆RB◆◆LB◆2\varepsilon_0◆RB◆$

$\square$

Connection to parallel plates. This result explains why the field between two large parallel plates is approximately uniform. Near the centre of the plates, each plate's contribution resembles that of an infinite plane. Between the plates, the fields add: $E_{\mathrm{total}} = \sigma/\varepsilon_0$. Outside the plates, the fields cancel: $E_{\mathrm{total}} = 0$. This is the ideal parallel-plate capacitor field, and it justifies the formula $E = V/d$ for the uniform field region.

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Board Coverage AQA: Gauss's law is not formally required but the uniform field result for parallel plates is given. Edexcel: Only the uniform field result $E = V/d$ is required. OCR (A): Gauss's law may appear as extension material in Module 6. CIE: The infinite plane result is not in the syllabus but is useful for understanding parallel plates.

3. Electric Potential

Definition. The electric potential $V$ at a point is the work done per unit positive charge to bring a small test charge from infinity to that point:

$\boxed{V = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r◆RB◆}$

SI units: volts (V), where 1 V = 1 J C$^{-1}$.

Derivation. The work done to bring $q$ from $\infty$ to $r$ against the electric force:

$W = \int_{\infty}^{r} F\,dr' = \int_{\infty}^{r} \frac◆LB◆Qq◆RB◆◆LB◆4\pi\varepsilon_0 r'^2◆RB◆\,dr' = \frac◆LB◆Qq◆RB◆◆LB◆4\pi\varepsilon_0◆RB◆\left[-\frac{1}{r'}\right]_{\infty}^{r} = \frac◆LB◆Qq◆RB◆◆LB◆4\pi\varepsilon_0 r◆RB◆$

$V = \frac{W}{q} = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r◆RB◆$

$\square$

Intuition. Potential is positive near a positive charge (work must be done against repulsion) and negative near a negative charge (work is done by the attraction). Potential decreases with distance, approaching zero at infinity.

Electric Potential Energy

The potential energy of two charges $q_1$ and $q_2$ separated by $r$:

$U = \frac◆LB◆q_1 q_2◆RB◆◆LB◆4\pi\varepsilon_0 r◆RB◆ = q_2 V_1$

where $V_1 = q_1/(4\pi\varepsilon_0 r)$ is the potential due to $q_1$ at the location of $q_2$.

Energy Stored in a Charged Capacitor

When a capacitor of capacitance $C$ carries charge $Q$ at p.d. $V$ ($Q = CV$), the energy stored is:

$\boxed{U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}}$

Proof of Energy in a Capacitor

The energy stored equals the total work done in transferring charge onto the capacitor. At any instant during charging, the p.d. across the plates is $v = q/C$. To transfer a small increment of charge $dq$ against this p.d., the work done is $dW = v\,dq = \frac{q}{C}\,dq$.

$W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{C}\left[\frac{q^2}{2}\right]_{0}^{Q} = \frac{Q^2}{2C}$

Since $Q = CV$: $U = \frac{1}{2}CV^2$. And since $V = Q/C$: $U = \frac{1}{2}QV$.

$\square$

Intuition. The factor of $\frac{1}{2}$ arises because the average p.d. during charging is $V/2$ (the p.d. grows linearly from 0 to $V$ as charge accumulates). The total work is therefore $W = Q \times V/2 = \frac{1}{2}QV$, which is the area of the triangle under the $V$--$q$ graph. This graphical interpretation is frequently tested: sketch the $V$--$Q$ graph and find the area to determine stored energy.

Where does the energy go? The energy is stored in the electric field between the plates. For a parallel-plate capacitor of area $A$ and separation $d$, the energy density (energy per unit volume) is:

$u = \frac{U}{Ad} = \frac◆LB◆\frac{1}{2}CV^2◆RB◆◆LB◆Ad◆RB◆ = \frac◆LB◆\frac{1}{2}\left(\frac◆LB◆\varepsilon_0 A◆RB◆◆LB◆d◆RB◆\right)\left(Ed\right)^2◆RB◆◆LB◆Ad◆RB◆ = \frac{1}{2}\varepsilon_0 E^2$

This result is general: any electric field of strength $E$ stores energy density $\frac{1}{2}\varepsilon_0 E^2$. Doubling the field strength quadruples the energy density.

Comparison with gravitational energy. Just as gravitational potential energy can be visualised as the area under the force--distance curve ($W = \int F\,dr$), the energy in a capacitor is the area under the voltage--charge curve. Both represent stored field energy, and both can be recovered as useful work when the system returns to its equilibrium state.

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info Topic 11 (capacitor energy and charging/discharging graphs). OCR (A): Required in 6.3.2, including deriving energy from the area under a $V$--$Q$ graph. CIE: Required in P3 (Chapter 22), including the three equivalent forms.

4. Relationship Between $E$ and $V$: $E = -dV/dr$

The electric field is the negative gradient of the electric potential.

Proof

Consider a test charge $q$ moved by a small distance $dr$ in the direction of the field. The work done by the field is:

$dW = F\,dr = qE\,dr$

This work comes from the change in potential energy: $dW = -q\,dV$ (negative because the field does work, reducing potential energy).

$qE\,dr = -q\,dV$

$\boxed{E = -\frac{dV}{dr}}$

In 3D: $\mathbf{E} = -\nabla V$.

Verification for a point charge. $V = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r◆RB◆$.

$-\frac{dV}{dr} = -\frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0◆RB◆\left(-\frac{1}{r^2}\right) = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r^2◆RB◆ = E \quad \checkmark$

Intuition. The minus sign tells us that the electric field points in the direction of decreasing potential — a positive charge "rolls downhill" in the potential landscape.

5. Charged Particle Motion in a Uniform Electric Field

Consider a particle of charge $q$ and mass $m$ entering a uniform electric field $E$ between parallel plates of length $L$, with initial velocity $v$ perpendicular to the field.

Horizontal motion (perpendicular to field): uniform velocity.

$x = vt, \qquad t = \frac{L}{v}$

Vertical motion (parallel to field): uniform acceleration.

$F = qE, \qquad a = \frac{qE}{m}$

$y = \frac{1}{2}at^2 = \frac{qE}{2m} \cdot \frac{L^2}{v^2}$

Derivation of the parabolic trajectory. Eliminating $t$: $y = \frac{qE}{2mv^2}x^2$. This is a parabola ($y \propto x^2$).

Vertical velocity at exit:

$v_y = at = \frac{qEL}{mv}$

Deflection angle:

$\tan\theta = \frac{v_y}{v} = \frac{qEL}{mv^2}$

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Example: Electron Deflection

An electron enters a uniform electric field of 5000 V m$^{-1}$ between plates of length 5.0 cm with speed $3.0 \times 10^7$ m s$^{-1}$. Calculate the vertical deflection.

Answer. $a = \frac{eE}{m_e} = \frac◆LB◆1.60 \times 10^{-19} \times 5000◆RB◆◆LB◆9.11 \times 10^{-31}◆RB◆ = 8.78 \times 10^{14}$ m s$^{-2}$.

$t = L/v = 0.050/3.0 \times 10^7 = 1.67 \times 10^{-9}$ s.

$y = \frac{1}{2}at^2 = \frac{1}{2} \times 8.78 \times 10^{14} \times (1.67 \times 10^{-9})^2 = 1.22 \times 10^{-3}$ m $= 1.22$ mm.

6. Real-World Applications of Electric Fields

Van de Graaff Generator

A Van de Graaff generator uses a moving rubber or latex belt to transport charge to a hollow metal dome. As charge accumulates on the outer surface of the dome, the electric potential rises (since $V = kQ/r$ and the dome has a large but finite radius). Potentials of several million volts can be achieved in laboratory demonstrations.

Key physics:

• Charge resides entirely on the outer surface of the dome. A conductor in electrostatic equilibrium has zero electric field inside, so no charge is needed on the inner surface to shield the interior.
• The dome is approximately spherical, so the field just outside is $E = kQ/r^2$, directed radially outward for a positively charged dome.
• When the field at the surface exceeds the breakdown strength of air ($E \approx 3 \times 10^6$ V m$^{-1}$ at STP), spark discharge occurs to any nearby grounded object.
• The potential at breakdown is $V_{\mathrm{breakdown}} = E_{\mathrm{breakdown}} \times r$. Larger domes reach higher breakdown potentials because the same surface charge density produces a weaker field on a larger sphere ($E = \sigma/\varepsilon_0$ for a sphere).

Safety demonstration. A person touching the dome while standing on an insulating platform also acquires a high potential. Their hair stands on end because each strand acquires the same sign of charge, and strands repel each other via Coulomb's law — a vivid demonstration of electrostatic repulsion.

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Board Coverage AQA: Van de Graaff generators are a specified practical context (required practical 10). Edexcel: Mentioned in CP3 and CP6 as an example of charge accumulation and high potential. OCR (A): Can appear in Paper 2 as a context question. CIE: Not explicitly named but the underlying physics (field of a sphere, potential, discharge) is examined in P4.

Inkjet Printers

Inkjet printers use electric fields to precisely control the trajectory of charged ink droplets. There are two main types:

1. Continuous inkjet (CIJ): A stream of ink is broken into uniform droplets by a vibrating piezoelectric crystal at a rate of roughly 100,000 droplets per second. Droplets pass through a charging electrode where each droplet is given a controlled charge. The charged droplets then pass between deflection plates creating a transverse electric field. The vertical deflection is $y = qEL^2/(2mv^2)$, so different charges produce different deflections. Uncharged droplets are collected in a gutter and recycled; charged droplets strike the paper to form characters.

2. Drop-on-demand (DOD): Thermal or piezoelectric actuators fire individual droplets only where needed. Some designs still use electric fields for final trajectory correction.

Physics involved. The deflection of each droplet in the plates is exactly the parabolic trajectory problem from Section 5. By varying the charge $q$ on each droplet (controlled by the charging electrode voltage), the printer achieves different vertical positions on the paper. The horizontal position is controlled by the timing of droplet ejection combined with paper motion.

Electrostatic Precipitator

Electrostatic precipitators remove fine particulate matter (dust, ash, smoke) from industrial exhaust gases. They are widely used in coal-fired power stations, waste incinerators, and cement plants, and can remove over 99% of particulate emissions.

How it works:

1. A thin high-voltage wire (at $\sim 50$ kV) runs along the centre of a duct or between large grounded collecting plates. The strong field near the wire ionises the surrounding air, creating a corona discharge.
2. The electric field (directed radially inward from the outer collecting plates towards the central wire, since the wire is negative) drives negative ions towards the collecting plates.
3. Particulate matter in the gas stream collides with these ions, acquires a negative charge, and is then attracted to the grounded collecting plates by the electric field.
4. The plates are periodically rapped (mechanically vibrated) to dislodge the collected dust into hoppers below for disposal or recycling.

Efficiency. Modern electrostatic precipitators can capture particles as small as $0.01\,\mu$m. The collection efficiency depends on the drift velocity of charged particles, which increases with both the charge acquired by the particle and the electric field strength. The Deutsch--Anderson equation relates collection efficiency $\eta$ to plate area, gas flow rate, and drift velocity.

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Board Coverage AQA: Electrostatic precipitators appear as an application of electric fields in Paper 2. Edexcel: Listed as an application in CP3. OCR (A): May appear as an unseen context question in Paper 2 requiring application of field and force principles. CIE: Industrial applications of electric fields are examinable in P4.

Problem Set

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Problem 1

Two point charges, $q_1 = +3.0\,\mu$C and $q_2 = -5.0\,\mu$C, are separated by 0.20 m. Calculate the electrostatic force between them.

Answer. $F = \frac{kq_1 q_2}{r^2} = \frac◆LB◆8.99 \times 10^9 \times 3.0 \times 10^{-6} \times 5.0 \times 10^{-6}◆RB◆◆LB◆0.04◆RB◆ = \frac{0.135}{0.04} = 3.37$ N (attractive).

If you get this wrong, revise: Coulomb's Law

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Problem 2

Calculate the electric field strength at a distance of 0.10 m from a point charge of $+8.0\,\mu$C.

Answer. $E = \frac{kQ}{r^2} = \frac◆LB◆8.99 \times 10^9 \times 8.0 \times 10^{-6}◆RB◆◆LB◆0.01◆RB◆ = \frac{71920}{0.01} = 7.19 \times 10^6$ N C$^{-1}$.

If you get this wrong, revise: Field of a Point Charge

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Problem 3

Calculate the electric potential at a distance of 5.0 cm from a $+2.0\,\mu$C point charge.

Answer. $V = \frac{kQ}{r} = \frac◆LB◆8.99 \times 10^9 \times 2.0 \times 10^{-6}◆RB◆◆LB◆0.050◆RB◆ = \frac{17980}{0.050} = 3.60 \times 10^5$ V $= 360$ kV.

If you get this wrong, revise: Electric Potential

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Problem 4

Starting from $E = -dV/dr$, derive the electric field of a point charge from its potential.

Answer. $V = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r◆RB◆$. $E = -\frac{dV}{dr} = -\frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0◆RB◆ \cdot \frac{d}{dr}\left(\frac{1}{r}\right) = -\frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0◆RB◆\left(-\frac{1}{r^2}\right) = \frac◆LB◆Q◆RB◆◆LB◆4\pi\varepsilon_0 r^2◆RB◆$.

If you get this wrong, revise: Relationship Between $E$ and $V$

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Problem 5

Two parallel plates are separated by 2.0 cm with a p.d. of 500 V across them. Calculate the electric field strength and the force on a proton between the plates.

Answer. $E = V/d = 500/0.020 = 25000$ V m$^{-1}$.

$F = qE = 1.60 \times 10^{-19} \times 25000 = 4.0 \times 10^{-15}$ N.

If you get this wrong, revise: Uniform Electric Field

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Problem 6

A proton is released from rest in a uniform electric field of $3.0 \times 10^4$ V m$^{-1}$. Calculate its acceleration and the kinetic energy gained after moving 5.0 cm.

Answer. $a = \frac{qE}{m} = \frac◆LB◆1.60 \times 10^{-19} \times 3.0 \times 10^4◆RB◆◆LB◆1.67 \times 10^{-27}◆RB◆ = 2.88 \times 10^{12}$ m s$^{-2}$.

$E_k = qV = qEd = 1.60 \times 10^{-19} \times 3.0 \times 10^4 \times 0.050 = 2.4 \times 10^{-16}$ J.

If you get this wrong, revise: Charged Particle Motion in a Uniform Electric Field

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Problem 7

Calculate the electric potential energy of two charges, $q_1 = +4.0\,\mu$C and $q_2 = +6.0\,\mu$C, separated by 0.30 m.

Answer. $U = \frac{kq_1 q_2}{r} = \frac◆LB◆8.99 \times 10^9 \times 4.0 \times 10^{-6} \times 6.0 \times 10^{-6}◆RB◆◆LB◆0.30◆RB◆ = \frac{0.2158}{0.30} = 0.719$ J.

If you get this wrong, revise: Electric Potential Energy

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Problem 8

An electron with speed $2.0 \times 10^7$ m s$^{-1}$ enters a uniform electric field of 8000 V m$^{-1}$ perpendicular to its motion. The plates are 8.0 cm long. Calculate the deflection angle.

Answer. $a = \frac{eE}{m_e} = \frac◆LB◆1.60 \times 10^{-19} \times 8000◆RB◆◆LB◆9.11 \times 10^{-31}◆RB◆ = 1.405 \times 10^{15}$ m s$^{-2}$.

$t = L/v = 0.080/(2.0 \times 10^7) = 4.0 \times 10^{-9}$ s.

$v_y = at = 1.405 \times 10^{15} \times 4.0 \times 10^{-9} = 5.62 \times 10^6$ m s$^{-1}$.

$\tan\theta = v_y/v = 5.62 \times 10^6 / 2.0 \times 10^7 = 0.281$. $\theta = 15.7^\circ$.

If you get this wrong, revise: Charged Particle Motion in a Uniform Electric Field

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Problem 9

Three equal charges of $+2.0\,\mu$C are placed at the corners of an equilateral triangle of side 0.10 m. Calculate the net electric field at the centre of the triangle.

Answer. The centre is at distance $r = \frac◆LB◆0.10◆RB◆◆LB◆\sqrt{3}◆RB◆ = 0.0577$ m from each charge.

$E_{\mathrm{one}} = \frac{kQ}{r^2} = \frac◆LB◆8.99 \times 10^9 \times 2.0 \times 10^{-6}◆RB◆◆LB◆0.00333◆RB◆ = 5.40 \times 10^6$ N C$^{-1}$.

By symmetry, the fields from the three charges cancel out (they are 120° apart and equal in magnitude). $E_{\mathrm{net}} = 0$.

If you get this wrong, revise: Field of a Point Charge

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Problem 10

A small charged sphere of mass 0.50 g is suspended by a thread in a uniform horizontal electric field of $5.0 \times 10^3$ V m$^{-1}$. The thread makes an angle of $15^\circ$ with the vertical. Calculate the charge on the sphere.

Answer. Resolving horizontally: $qE = T\sin 15^\circ$. Resolving vertically: $mg = T\cos 15^\circ$.

$\tan 15° = \frac{qE}{mg}$. $q = \frac◆LB◆mg\tan 15°◆RB◆◆LB◆E◆RB◆ = \frac◆LB◆0.50 \times 10^{-3} \times 9.81 \times 0.2679◆RB◆◆LB◆5000◆RB◆ = \frac◆LB◆1.314 \times 10^{-3}◆RB◆◆LB◆5000◆RB◆ = 2.63 \times 10^{-7}$ C $= 263$ nC.

If you get this wrong, revise: Electric Field Strength

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Problem 11

A Van de Graaff generator has a dome of radius 0.30 m. Calculate the electric field strength and the potential at the surface when the dome carries a charge of $1.0\,\mu$C. At what potential will spark discharge occur if the breakdown field of air is $3.0 \times 10^6$ V m$^{-1}$?

Answer. $E = \frac{kQ}{r^2} = \frac◆LB◆8.99 \times 10^9 \times 1.0 \times 10^{-6}◆RB◆◆LB◆0.090◆RB◆ = 9.99 \times 10^4$ V m$^{-1}$.

$V = \frac{kQ}{r} = \frac◆LB◆8.99 \times 10^9 \times 1.0 \times 10^{-6}◆RB◆◆LB◆0.30◆RB◆ = 3.00 \times 10^4$ V $= 30.0$ kV.

For breakdown: $V_{\mathrm{breakdown}} = E_{\mathrm{breakdown}} \times r = 3.0 \times 10^6 \times 0.30 = 9.0 \times 10^5$ V $= 900$ kV.

If you get this wrong, revise: Field of a Point Charge, Electric Potential, Van de Graaff Generator

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Problem 12

A $10$ nF capacitor is charged to $200$ V. Calculate the energy stored. The capacitor is then disconnected from the supply and the plate separation is doubled. Calculate the new energy stored and explain where the extra energy came from.

Answer. Initial energy: $U = \frac{1}{2}CV^2 = \frac{1}{2} \times 10 \times 10^{-9} \times (200)^2 = 2.0 \times 10^{-4}$ J $= 200\,\mu$J.

Since the capacitor is disconnected, charge $Q$ is conserved. When $d$ doubles, $C' = C/2 = 5.0$ nF.

$V' = Q/C' = (CV)/C' = (10 \times 10^{-9} \times 200)/(5.0 \times 10^{-9}) = 400$ V.

$U' = \frac{1}{2}C'V'^2 = \frac{1}{2} \times 5.0 \times 10^{-9} \times (400)^2 = 4.0 \times 10^{-4}$ J $= 400\,\mu$J.

The energy doubled because work was done against the attractive force between the oppositely charged plates when pulling them apart. This mechanical work is stored as additional electric field energy.

If you get this wrong, revise: Energy Stored in a Charged Capacitor

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Problem 13

In an electrostatic precipitator, a dust particle of mass $1.0 \times 10^{-12}$ kg acquires a charge of $-50$ fC and enters a region with electric field $2.0 \times 10^5$ V m$^{-1}$ directed towards the collecting plate. If the particle must travel $0.50$ m to reach the plate, how long does it take? (Assume uniform field and negligible air resistance.)

Answer. $F = qE = 50 \times 10^{-15} \times 2.0 \times 10^5 = 1.0 \times 10^{-8}$ N.

$a = F/m = 1.0 \times 10^{-8} / 1.0 \times 10^{-12} = 1.0 \times 10^4$ m s$^{-2}$.

$s = \frac{1}{2}at^2$, so $t = \sqrt{2s/a} = \sqrt◆LB◆2 \times 0.50 / 1.0 \times 10^4◆RB◆ = \sqrt◆LB◆1.0 \times 10^{-4}◆RB◆ = 0.010$ s $= 10$ ms.

If you get this wrong, revise: Charged Particle Motion in a Uniform Electric Field, Electrostatic Precipitator

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Problem 14

Two charges, $q_A = +4.0\,\mu$C and $q_B = -2.0\,\mu$C, are placed on the x-axis at $x = 0$ and $x = 0.20$ m respectively. Find the point on the x-axis (outside the two charges) where the electric field is zero.

Answer. Let the zero-field point be at distance $x$ from charge $A$ (to the right of charge $B$), so it is at distance $(x - 0.20)$ from charge $B$.

For the fields to cancel (both point rightwards at this location, since $A$ repels a positive test charge and $B$ attracts):

$\frac{kq_A}{x^2} = \frac{kq_B}{(x - 0.20)^2}$

$\frac{4.0}{x^2} = \frac{2.0}{(x - 0.20)^2}$

$2(x - 0.20)^2 = x^2$

$2x^2 - 0.80x + 0.08 = x^2$

$x^2 - 0.80x + 0.08 = 0$

$x = \frac◆LB◆0.80 \pm \sqrt{0.64 - 0.32}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆0.80 \pm 0.566◆RB◆◆LB◆2◆RB◆$

Taking the solution outside both charges: $x = \frac{0.80 + 0.566}{2} = 0.683$ m.

The zero-field point is at $x = 0.683$ m on the x-axis.

If you get this wrong, revise: Field of a Point Charge

Details

Problem 15

Compare the gravitational and electrostatic forces between a proton and an electron in a hydrogen atom (separation $r = 5.29 \times 10^{-11}$ m). By what factor does the electrostatic force exceed the gravitational force?

Answer. $F_e = \frac{kq_1q_2}{r^2} = \frac◆LB◆8.99 \times 10^9 \times (1.60 \times 10^{-19})^2◆RB◆◆LB◆(5.29 \times 10^{-11})^2◆RB◆ = \frac◆LB◆2.30 \times 10^{-28}◆RB◆◆LB◆2.80 \times 10^{-21}◆RB◆ = 8.22 \times 10^{-8}$ N.

$F_g = \frac{Gm_pm_e}{r^2} = \frac◆LB◆6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31}◆RB◆◆LB◆(5.29 \times 10^{-11})^2◆RB◆ = \frac◆LB◆1.01 \times 10^{-67}◆RB◆◆LB◆2.80 \times 10^{-21}◆RB◆ = 3.61 \times 10^{-47}$ N.

Ratio: $\frac{F_e}{F_g} = \frac◆LB◆8.22 \times 10^{-8}◆RB◆◆LB◆3.61 \times 10^{-47}◆RB◆ = 2.28 \times 10^{39}$.

The electrostatic force is $\sim 10^{39}$ times stronger than gravity. This enormous ratio explains why atomic and molecular physics are governed entirely by electromagnetic forces, with gravity only becoming significant at macroscopic scales where charges cancel almost perfectly.

If you get this wrong, revise: Coulomb's Law, Comparison with Gravity

Board-Specific Notes

AQA (7408)

• Electric fields are assessed in Paper 2 (Sections 6.1--6.2).
• Coulomb's law and the inverse square law formula are provided on the data sheet.
• Required practical 10 investigates electric fields using conducting paper (mapping equipotentials and field lines).
• The comparison with gravitational fields is a common synoptic link, particularly in multi-mark questions.
• Graphical methods for determining energy stored (area under $V$--$Q$ curve) are frequently examined.

Edexcel (9PH0)

• Electric fields are covered in Core Practical 3 (CP3) and Topic 11.
• Edexcel places emphasis on practical applications and numerical problem-solving using the formula booklet.
• The energy stored in capacitors is covered in Topic 11 alongside charging and discharging through resistors.
• Deflection of charged particles (the "electron gun" and CRT context) is a recurring exam theme.
• Questions often require identifying the correct equation from the formula sheet for multi-step calculations.

OCR (A) (H556)

• Electric fields appear in Module 6 (Sections 6.1 and 6.3).
• OCR commonly tests full derivations from first principles (e.g., deriving $E = -dV/dr$ or the potential of a point charge from Coulomb's law).
• The $V$--$Q$ graph area-under-curve argument for energy is a staple of OCR marking schemes.
• Gravitational--electrostatic comparison questions are a hallmark of OCR synoptic assessment.
• Multi-part questions often begin with a recall/derivation and build to an application or calculation.

CIE (9702)

• Electric fields are in Paper 4 (A2), Chapters 21--22 of the coursebook.
• CIE requires the most mathematical rigour: full derivations from Coulomb's law to potential and field strength are expected.
• CIE commonly combines electric fields with mechanics (projectile motion of charged particles in fields).
• Permittivity calculations and relative permittivity ($\varepsilon_r$) of dielectric materials are explicitly examinable.
• Numerical answers are typically given to 2 or 3 significant figures, matching the precision of the data provided.

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tip

Diagnostic Test Ready to test your understanding of Electric Fields? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Electric Fields with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

danger

Common Pitfalls

• Confusing electric field strength E with electric potential V: Electric field strength E is the FORCE per unit charge (a vector, measured in V/m or N/C). Electric potential V is the ENERGY per unit charge (a scalar, measured in volts). They are related by E = -dV/dr, but they are fundamentally different quantities.

• Forgetting that electric field lines point from positive to negative: Field lines show the direction of force on a POSITIVE test charge. If you place a negative charge in the field, the force acts in the OPPOSITE direction to the field lines. Students often draw force arrows in the wrong direction for negative charges.

• Not distinguishing between uniform and radial fields: For a parallel plate capacitor, E is uniform (constant everywhere between the plates) and V varies linearly with distance. For a point charge, E follows an inverse square law and V follows an inverse relationship (V proportional to 1/r). Using the wrong field model leads to incorrect answers.

• Assuming work done is always positive when moving a charge: Work done by the electric field is positive when a positive charge moves in the direction of the field (from high to low potential). Moving a charge AGAINST the field requires external work and the work done by the field is negative. Always consider the sign of the charge and the direction of movement.