Electric Fields

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

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

Where $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$ is the permittivity of free space and $k = 1/(4\pi\varepsilon_0) = 8.99 \times 10^9$ N m$^2$ C$^{-2}$ is Coulomb’s constant.

The force is repulsive for like charges and attractive for opposite charges, directed along the Line joining them.

The net force on a charge due to multiple other charges is the vector sum of the individual Coulomb forces:

$\mathbf{F}_{\mathrm{net}} = \sum_i \frac{q\,q_i}{4\pi\varepsilon_0 r_i^2}\,\hat{\mathbf{r}}_i$

This linearity is fundamental: each pair of charges interacts independently of all others.

Property	Gravitational	Electrostatic
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 approximately $10^{36}$ times stronger than gravity for proton—proton Interactions. This enormous ratio explains why atomic and molecular structure is governed entirely by Electromagnetic forces.

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

$\boxed{\mathbf{E} = \frac{\mathbf{F}}{q}}$

SI units: N C$^{-1}$Equivalent to V m$^{-1}$.

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

Proof. Place test charge $q$ at distance $r$ from $Q$. By Coulomb’s law: $F = Qq/(4\pi\varepsilon_0 r^2)$. Therefore $E = F/q = Q/(4\pi\varepsilon_0 r^2)$. $\square$

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

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

Where $V$ is the potential difference and $d$ is the plate separation.

Proof. A charge $q$ between the plates experiences force $F = qE$. Work done moving from one plate To the other: $W = Fd = qEd$. But also $W = qV$. Therefore $qEd = qV$Giving $E = V/d$. $\square$

The field is uniform (constant magnitude and direction) between the plates, with fringe effects at the Edges.

Field lines provide a visual representation of the electric field:

• The direction of the line at any point gives the direction of $\mathbf{E}$.
• The density of lines is proportional to the field strength.
• Lines begin on positive charges and end on negative charges.
• Lines never cross (the field has a unique direction at every point).
• Lines are perpendicular to conducting surfaces at equilibrium.

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

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

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

$V = \frac{W}{q} = \frac{1}{q}\int_{\infty}^{r}\frac{Qq}{4\pi\varepsilon_0 r'^2}\,dr' = \frac{Q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r} = \frac{Q}{4\pi\varepsilon_0 r}$

$\square$

Sign convention. Potential is positive near a positive charge (work must be done against repulsion) And negative near a negative charge (the field does work). Potential decreases with distance, Approaching zero at infinity.

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

Proof. Consider a test charge $q$ moved by $dr$ in the direction of the field. Work done by the Field: $dW = qE\,dr$. This equals the loss in potential energy: $dW = -q\,dV$. Therefore $qE\,dr = -q\,dV$Giving $E = -dV/dr$. $\square$

The minus sign means the field points in the direction of decreasing potential.

Verification for a point charge. $V = Q/(4\pi\varepsilon_0 r)$.

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

$\boxed{U = \frac{q_1 q_2}{4\pi\varepsilon_0 r}}$

This is the work required to bring two charges from infinite separation to distance $r$.

Definition. An equipotential surface is a surface on which every point has the same electric Potential.

1. No work is done moving a charge along an equipotential surface (since $\Delta V = 0$).
2. The electric field is always perpendicular to equipotential surfaces (since $E = -dV/dr$ and $dV = 0$ along the surface).
3. Equipotential surfaces never cross (each point has a unique potential).
4. For a point charge, equipotentials are concentric spheres.
5. For a uniform field, equipotentials are parallel planes perpendicular to the field.
6. Equipotentials are closer together where the field is stronger (steeper potential gradient).

A practical method uses conducting paper with electrodes painted on:

1. Connect electrodes to a power supply, establishing a potential difference.
2. Use a voltmeter probe to locate points of equal potential.
3. Plot the equipotential lines by joining points of equal voltage.
4. Draw field lines perpendicular to the equipotentials.

## 5. Motion of Charged Particles in Uniform Fields

A particle of charge $q$ and mass $m$ enters a uniform electric field $E$ with initial velocity $v$ Perpendicular to the field, between plates of length $L$.

Horizontal (perpendicular to field): uniform motion.

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

Vertical (parallel to field): uniformly accelerated.

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

$y = \frac{1}{2}at^2 = \frac{qEL^2}{2mv^2}$

Eliminating $t$: $y = \frac{qE}{2mv^2}\,x^2$. This is a parabola.

Vertical velocity at exit: $v_y = at = \frac{qEL}{mv}$.

Deflection angle: $\tan\theta = \frac{v_y}{v} = \frac{qEL}{mv^2}$.

An alternative approach uses energy conservation. The kinetic energy gained by the particle equals the Work done by the field:

$\Delta E_k = qV = qEd$

Where $d$ is the vertical displacement. This is often quicker than the kinematic approach.

Worked Example: Electron Deflection

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

Answer. $a = \frac{eE}{m_e} = \frac{1.60 \times 10^{-19} \times 5000}{9.11 \times 10^{-31}} = 8.78 \times 10^{14}$ 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.

$v_y = at = 8.78 \times 10^{14} \times 1.67 \times 10^{-9} = 1.47 \times 10^6$ m s$^{-1}$.

$\tan\theta = v_y/v = 1.47 \times 10^6 / (3.0 \times 10^7) = 0.0489$. $\theta = 2.80^\circ$.

A CRT uses electric fields to control and deflect a beam of electrons:

1. Electron gun: A heated cathode emits electrons by thermionic emission. A high potential difference $V_{\mathrm{acc}}$ accelerates them through a potential difference, giving kinetic energy $\frac{1}{2}m_e v^2 = eV_{\mathrm{acc}}$.
2. Deflection system: Two pairs of parallel plates (X and Y) apply transverse electric fields, deflecting the beam horizontally and vertically.
3. Fluorescent screen: Electrons strike a phosphor coating, producing visible light.

From energy conservation:

$\frac{1}{2}m_e v^2 = eV_{\mathrm{acc}}$

$\boxed{v = \sqrt{\frac{2eV_{\mathrm{acc}}}{m_e}}}$

For $V_{\mathrm{acc}} = 2000$ V: $v = \sqrt{2 \times 1.60 \times 10^{-19} \times 2000 / 9.11 \times 10^{-31}} = 2.65 \times 10^7$ m s$^{-1}$.

The deflection sensitivity $S$ is the deflection per unit deflection voltage:

$S = \frac{y}{V_d} = \frac{eL^2}{2m_e v^2 d} = \frac{L^2}{4V_{\mathrm{acc}}\,d}$

Where $L$ is the plate length and $d$ is the plate separation. Higher sensitivity requires longer Plates, closer spacing, and lower acceleration voltage.

## 7. Electric Fields of Extended Charge Distributions

A ring of radius $a$ carrying total charge $Q$. The field at distance $x$ from the centre along the Axis:

$\boxed{E = \frac{Qx}{4\pi\varepsilon_0(x^2 + a^2)^{3/2}}}$

Proof. By symmetry, the transverse components cancel. Each element $dq$ contributes $dE = dq/(4\pi\varepsilon_0(x^2 + a^2))$. The axial component is $dE_x = dE \cdot x/\sqrt{x^2 + a^2}$. Integrating over the ring:

$E_x = \frac{Q}{4\pi\varepsilon_0}\cdot\frac{x}{(x^2 + a^2)^{3/2}}$

$\square$

Checks. At $x = 0$: $E = 0$ (by symmetry). For $x \gg a$: $E \approx Q/(4\pi\varepsilon_0 x^2)$ (point charge limit). $\checkmark$

For a line of charge with linear charge density $\lambda$ (C m$^{-1}$):

$\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$

Where $r$ is the perpendicular distance from the line. Note: the field falls off as $1/r$Not $1/r^2$ Because a line charge is an extended source in one dimension.

Millikan (1909—1913) measured the elementary charge $e$ by observing electrically charged oil drops In a uniform electric field.

Method: An oil drop of mass $m$ carries charge $q$. In a uniform upward field $E$The drop is Suspended when the electric force balances gravity:

$qE = mg$

$\boxed{q = \frac{mg}{E}}$

The mass is found from the terminal velocity (using Stokes’ law for the drag force in air). Millikan Found that all measured charges were integer multiples of $e = 1.60 \times 10^{-19}$ C.

Significance. This experiment proved that charge is quantised — it comes in discrete packets of Size $e$.

For a parallel-plate capacitor with plate area $A$ and separation $d$:

$C = \frac{\varepsilon_0 A}{d}$

The energy stored when the capacitor carries charge $Q$ at potential difference $V$:

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

Proof. During charging, the p.d. At any instant is $v = q/C$. Work to transfer charge $dq$: $dW = v\,dq = q\,dq/C$.

$W = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}$

$\square$

The energy is stored in the electric field between the plates. The energy density is:

$u = \frac{1}{2}\varepsilon_0 E^2$

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 Force between them.

Answer. $F = \frac{k|q_1||q_2|}{r^2} = \frac{8.99 \times 10^9 \times 3.0 \times 10^{-6} \times 5.0 \times 10^{-6}}{0.040} = 3.37$ N (attractive).

Problem 2

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

Answer. $E = \frac{kQ}{r^2} = \frac{8.99 \times 10^9 \times 8.0 \times 10^{-6}}{0.010} = 7.19 \times 10^6$ N C$^{-1}$.

Problem 3

Two parallel plates are separated by 2.0 cm with p.d. 500 V. Calculate the field strength and the force On a proton between the plates.

Answer. $E = V/d = 500/0.020 = 2.5 \times 10^4$ V m$^{-1}$. $F = qE = 1.60 \times 10^{-19} \times 2.5 \times 10^4 = 4.0 \times 10^{-15}$ N.

Problem 4

Calculate the electric potential at 5.0 cm from a $+2.0\,\mu$C point charge. A second charge of $-1.0\,\mu$C is placed at this point. Calculate the potential energy of the system.

Answer. $V = \frac{kQ}{r} = \frac{8.99 \times 10^9 \times 2.0 \times 10^{-6}}{0.050} = 3.60 \times 10^5$ V.

$U = q_2 V = (-1.0 \times 10^{-6})(3.60 \times 10^5) = -0.360$ J.

Problem 5

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

Answer. $V = Q/(4\pi\varepsilon_0 r)$. $E = -\frac{dV}{dr} = -\frac{Q}{4\pi\varepsilon_0}\cdot\frac{d}{dr}(r^{-1}) = -\frac{Q}{4\pi\varepsilon_0}(-r^{-2}) = \frac{Q}{4\pi\varepsilon_0 r^2}$. $\square$

Problem 6

An electron is accelerated through 3000 V in a CRT. Calculate its final speed and kinetic energy.

Answer. $E_k = eV = 1.60 \times 10^{-19} \times 3000 = 4.80 \times 10^{-16}$ J.

$v = \sqrt{2E_k/m_e} = \sqrt{2 \times 4.80 \times 10^{-16}/9.11 \times 10^{-31}} = 3.25 \times 10^7$ m s$^{-1}$.

Problem 7

In a Millikan-type experiment, an oil drop of mass $1.2 \times 10^{-14}$ kg is suspended between Parallel plates with field $E = 4.8 \times 10^4$ V m$^{-1}$. Calculate the charge on the drop and Determine how many elementary charges it carries.

Answer. $q = mg/E = 1.2 \times 10^{-14} \times 9.81 / (4.8 \times 10^4) = 2.45 \times 10^{-18}$ C.

$n = q/e = 2.45 \times 10^{-18} / 1.60 \times 10^{-19} = 15.3$.

Since $n$ must be an integer, the drop carries 15 elementary charges (the discrepancy is within Experimental uncertainty).

Problem 8

Sketch the equipotential lines and field lines for two equal positive charges separated by distance $d$. Explain why the field is zero at the midpoint.

Answer. The equipotential lines form peanut-shaped closed curves around each charge, with a Zero-potential surface at infinity. The field lines radiate outward from each charge, curving away from Each other.

At the midpoint, the fields due to each charge are equal in magnitude ($kq/(d/2)^2$) and opposite in Direction (each points away from its own charge). By symmetry, $\mathbf{E}_1 + \mathbf{E}_2 = 0$. This is an unstable equilibrium point.

Problem 9

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. $a = qE/m_p = 1.60 \times 10^{-19} \times 3.0 \times 10^4 / 1.67 \times 10^{-27} = 2.88 \times 10^{12}$ m s$^{-2}$.

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

Problem 10

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

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

$\tan 15^\circ = qE/(mg)$. $q = mg\tan 15^\circ / E = 0.50 \times 10^{-3} \times 9.81 \times 0.268 / 5000 = 2.63 \times 10^{-7}$ C $= 263$ nC.

1. Confusing EMF and potential difference. EMF is the total energy per unit charge supplied; PD is the energy per unit charge transferred to a component.

2. Forgetting that ammeters are connected in series and voltmeters in parallel.

3. Incorrectly applying Kirchhoff’s second law by missing components in a loop.

4. Confusing scalar and vector quantities. Always check whether direction matters for the quantity in question.

5. Misidentifying the system boundary when applying conservation laws. Define what is included before writing equations.

6. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

Problem. Two parallel plates are separated by $5\ \mathrm{cm}$ and have a potential difference of $2000\ \mathrm{V}$ across them. Find the electric field strength and the force on an electron between the plates.

Solution. $E = \frac{V}{d} = \frac{2000}{0.05} = 40000\ \mathrm{V\,m^{-1}}$

Force on the electron:

$F = eE = 1.6 \times 10^{-19} \times 40000 = 6.4 \times 10^{-15}\ \mathrm{N}$

$\blacksquare$

Problem. Two point charges $+3\ \mathrm{nC}$ and $-5\ \mathrm{nC}$ are separated by $10\ \mathrm{cm}$ in vacuum. Calculate the electrostatic force between them.

Solution. $F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2} = \frac{8.99 \times 10^9 \times 3 \times 10^{-9} \times 5 \times 10^{-9}}{(0.1)^2}$

$F = \frac{8.99 \times 10^9 \times 15 \times 10^{-18}}{0.01} = \frac{134.85 \times 10^{-9}}{0.01} = 1.35 \times 10^{-5}\ \mathrm{N}$

The force is attractive (opposite charges).

$\blacksquare$

• Coulomb’s law: $F = \frac{q_1 q_2}{4\pi\varepsilon_0 r^2}$; like charges repel, unlike attract.
• Electric field strength: $E = \frac{F}{q}$; for a point charge $E = \frac{Q}{4\pi\varepsilon_0 r^2}$.
• Uniform field between parallel plates: $E = \frac{V}{d}$; equipotential lines are perpendicular to field lines.
• Electric potential: $V = \frac{Q}{4\pi\varepsilon_0 r}$; potential energy: $W = qV$.
• $E = -\frac{dV}{dr}$; the electric field is the negative gradient of the potential.

The universe operates through fundamental forces and energy transfers. Forces are pushes and pulls that change motion, energy is the currency that drives all processes, and waves transfer energy without transferring matter. These principles connect seemingly different phenomena - from the orbit of planets to the vibration of atoms - under unified explanations that reveal the elegant simplicity underlying nature’s complexity.

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