Electric Field and Field Strength

Coulomb's law tells you the force between two charges, but it says nothing until a second charge is present. The electric field is the device that removes this awkwardness: it assigns, to every point in space around a charge, a vector that says what force would act on a test charge placed there. The field exists whether or not anything is there to feel it, and it turns "action at a distance" into a local statement — a charge responds to the field at its own location, not directly to a charge somewhere far away.

This chapter defines field strength, works out the field of a point charge and of a uniform arrangement, and introduces the two pictures used to visualise fields: field lines and, for the uniform case, the link between field and potential difference. The uniform field between parallel plates, derived here, is the setting for almost every numerical electric-field problem in the rest of the block, and for the particle-deflection experiments of Chapters 8 and 13.

Reading note. The relation $E = \dfrac{V}{d}$ for a uniform field is used constantly and is worth committing to memory now, along with its two unit forms: $\dfrac{\text{N}}{\text{C}}$ and $\dfrac{\text{V}}{\text{m}}$ are the same thing.

1. Definition of Electric Field Strength

Definition. The electric field strength $\vec{E}$ at a point is the force per unit positive charge experienced by a small stationary test charge placed at that point:

$$\vec{E} = \frac{\vec{F}}{q}$$

The unit is the newton per coulomb, $\dfrac{\text{N}}{\text{C}}$. $\vec{E}$ is a vector, pointing in the direction of the force on a positive charge; a negative charge in the same field feels a force in the opposite direction to $\vec{E}$.

Two cautions are built into the word "small". The test charge must be small enough in magnitude that its own field does not disturb the charges producing $\vec{E}$, and it is taken to be a point so that "the field at a point" is well defined. Formally, $\vec{E} = \lim_{q\to 0}\vec{F}/q$.

Once $\vec{E}$ is known at a point, the force on any charge $q$ placed there is simply

This is the equation that couples a charge to a field, and it reappears in every subsequent chapter.

2. The Field of a Point Charge

Put a test charge $q$ a distance $r$ from a point charge $Q$. Coulomb's law gives the force magnitude $F = k\dfrac{|Q||q|}{r^{2}}$, so dividing by $|q|$:

The field of a point charge is radial: it points directly away from $Q$ if $Q$ is positive, and directly towards $Q$ if $Q$ is negative. Its magnitude falls off as $\dfrac{1}{r^{2}}$, exactly like the force — because the test charge is a fixed number, dividing by it does not change the distance dependence.

For several source charges, superposition carries over from the force: the total field is the vector sum of the individual point-charge fields,

Worked example — field on the perpendicular bisector of a dipole. Charges $+Q$ and $-Q$ are separated by $2a$. Find the field at a point on the perpendicular bisector, a distance $y$ from the centre, for $y \gg a$.

Each charge is a distance $\sqrt{y^{2}+a^{2}}$ from the point and contributes a field of magnitude $E_1 = k\dfrac{Q}{y^{2}+a^{2}}$. The vertical components cancel; the horizontal components (pointing from $+Q$ towards $-Q$) add. Each horizontal component is $E_1\cos\theta$ with $\cos\theta = \dfrac{a}{\sqrt{y^{2}+a^{2}}}$, so

The dipole field dies as $\dfrac{1}{r^{3}}$, faster than a single charge, because the two charges nearly cancel at large distance.

3. Field Lines

A field line is a directed curve whose tangent at every point gives the direction of $\vec{E}$ there. The rules for drawing them:

• Lines begin on positive charge and end on negative charge (or run to or from infinity).
• The number of lines leaving or entering a charge is proportional to the magnitude of the charge.
• Lines never cross — the field has one direction at each point.
• Where lines are closer together, the field is stronger; evenly spaced parallel lines mean a uniform field.
• Field lines meet the surface of a conductor at right angles, because any component along the surface would drive a current until it vanished.

For a single positive point charge the lines are radial spokes. For two equal opposite charges (a dipole) they curve from $+$ to $-$. For two equal positive charges there is a neutral point midway between them where the two fields cancel and $E = 0$.

4. The Uniform Field Between Parallel Plates

Two large parallel conducting plates a distance $d$ apart, carrying equal and opposite charge, produce a field that is uniform in the region between them (edge effects aside): constant in magnitude and direction, represented by equally spaced parallel lines running from the positive to the negative plate.

If a potential difference $V$ is applied across the plates, the work done moving a charge $q$ from one plate to the other is $W = qV$. But that same work, done by the constant electric force $F = qE$ over the distance $d$, is $W = qEd$. Equating:

Uniform-field result. Between parallel plates a distance $d$ apart with potential difference $V$,

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

directed from the positive plate to the negative plate. The unit $\dfrac{\text{V}}{\text{m}}$ is identical to $\dfrac{\text{N}}{\text{C}}$.

A charged particle released in this field feels a constant force $qE$ and therefore a constant acceleration $a = \dfrac{qE}{m}$ — the motion is exactly the uniformly accelerated motion of Block A, and if the particle enters sideways it follows a parabola, just like a projectile under gravity.

5. Worked Example: Deflecting an Electron

An electron enters midway between two horizontal plates $4.0\ \text{cm}$ long and $1.2\ \text{cm}$ apart, moving horizontally at $v_x = 2.0\times 10^{7}\ \dfrac{\text{m}}{\text{s}}$. The plates carry a potential difference of $180\ \text{V}$. Find the vertical deflection as the electron leaves the plates. Take $m_e = 9.11\times 10^{-31}\ \text{kg}$, $e = 1.60\times 10^{-19}\ \text{C}$.

Field between the plates:

Vertical acceleration:

Time in the field: $t = \dfrac{L}{v_x} = \dfrac{0.040}{2.0\times 10^{7}} = 2.0\times 10^{-9}\ \text{s}$.

Vertical deflection, from $y = \tfrac{1}{2}at^{2}$:

Since the electron started $6.0\ \text{mm}$ from each plate, it just clears the plates. This is the principle of the cathode-ray tube.

6. Field Near a Charged Conductor

In electrostatic equilibrium, the field inside a conductor is zero: if it were not, free charges would move until it was. Consequences:

• Any excess charge on an isolated conductor sits entirely on its outer surface.
• The field just outside the surface is perpendicular to it.
• Charge concentrates where the surface curves most sharply, so the field is strongest at points and edges — the basis of the lightning conductor.
• A hollow conductor shields its interior from external fields (a Faraday cage): the cavity field is zero regardless of the charge outside.

Common Exam Pitfalls

• Confusing $\vec{E}$ with $\vec{F}$. Field strength is force per unit charge; it exists at a point whether or not a charge is there. Multiply by $q$ to get the force, and remember a negative $q$ reverses the direction.
• Forgetting that the field of a negative charge points inward. $E = k|Q|/r^{2}$ gives only the magnitude; the direction must be assigned from the sign of $Q$.
• Adding fields as scalars. Like forces, fields superpose as vectors — resolve into components before summing, and watch for neutral points where they cancel.
• Using $E = V/d$ for a non-uniform field. It is exact only for a uniform field (parallel plates). The field of a point charge needs $E = kQ/r^{2}$.
• Mixing up which plate is positive. The field, and the force on a positive charge, point from the positive plate to the negative plate.
• Leaving distances in centimetres. $d$ in $E = V/d$ must be in metres, or the field comes out a factor of 100 wrong.

Summary

• Electric field strength: $\vec{E} = \dfrac{\vec{F}}{q}$, a vector in $\dfrac{\text{N}}{\text{C}}$ pointing along the force on a positive charge. The force on any charge is $\vec{F} = q\vec{E}$.
• Point charge: $E = k\dfrac{|Q|}{r^{2}}$, radial — outward from positive $Q$, inward to negative $Q$. Fields superpose as vectors.
• Field lines run from positive to negative charge, never cross, and are denser where the field is stronger; parallel evenly spaced lines mean a uniform field.
• Parallel plates: $E = \dfrac{V}{d}$, uniform, directed from $+$ to $-$ plate; $\dfrac{\text{V}}{\text{m}} \equiv \dfrac{\text{N}}{\text{C}}$. A charge in this field has constant acceleration $a = \dfrac{qE}{m}$.
• Inside a conductor in equilibrium $E = 0$; excess charge lies on the outer surface, the external field meets the surface at right angles and is largest at sharp points.