Electric Field

Why the field is introduced

If a charge $Q$ at the origin exerts a force on a charge $q$ at $P$, and we then remove $q$, what is left at $P$? The field concept answers this: $Q$ produces an electric field everywhere in the space around it, and when $q$ is brought to $P$, it is the field there that acts on it and produces the force. The interaction is between the charge and the field, not action at a distance.

Definition

$$\vec{E} = \lim_{q_0\to0}\dfrac{\vec{F}}{q_0}$$

The electric field at a point is the force that would be experienced by a unit positive test charge placed at that point. The limit $q_0\to0$ is written because the test charge would otherwise disturb the source charge $Q$; making $q_0$ negligibly small makes $\vec{F}$ negligibly small but keeps the ratio $\vec{F}/q_0$ finite.

• The charge producing the field is the source charge, the charge used to detect it is the test charge.
• The force on any charge $q$ placed in a field is $\vec{F} = q\vec{E}$. This equation defines the SI unit of field, $\text{N C}^{-1}$; an alternative unit $\text{V m}^{-1}$ appears in Chapter 2.
• On a positive charge the force is along $\vec{E}$; on a negative charge it is opposite to $\vec{E}$, since $\vec{F} = -\lvert q\rvert\vec{E}$. A negative charge released in a field therefore moves opposite to $\vec{E}$, not along it.

Field of a point charge

$$\vec{E}(\vec{r}) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\,\hat{r}$$

where $\hat{r}$ is the unit vector from the charge to the field point.

• Direction: radially outward for a positive source charge, radially inward for a negative one.
• Magnitude depends only on $r$, so it is the same at every point of a sphere centred on the charge — the field has spherical symmetry.
• $\vec{E}$ is independent of the test charge: since $\vec{F}\propto q_0$, the ratio $\vec{F}/q_0$ does not involve $q_0$ at all. It does depend on position, and exists at every point of three-dimensional space.

Field of a system of charges — the 5-mark derivation

Consider $n$ charges $q_1, q_2, \ldots, q_n$ with position vectors $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n$ relative to an origin $O$, and a point $P$ with position vector $\vec{r}_P$.

By Coulomb's law, the force on a test charge $q_0$ at $P$ due to $q_1$ alone is

$$\vec{F}_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_0}{r_{1P}^{2}}\,\hat{r}_{1P}$$

where $r_{1P}$ is the distance from $q_1$ to $P$ and $\hat{r}_{1P}$ the unit vector from $q_1$ towards $P$. Dividing by $q_0$, the field at $P$ due to $q_1$ is

$$\vec{E}_1 = \dfrac{\vec{F}_1}{q_0} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}^{2}}\,\hat{r}_{1P}$$

and similarly $\vec{E}_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_2}{r_{2P}^{2}}\hat{r}_{2P}$, and so on. By the principle of superposition,

$$\vec{E} = \vec{E}_1+\vec{E}_2+\cdots+\vec{E}_n = \dfrac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\dfrac{q_i}{r_{iP}^{2}}\,\hat{r}_{iP}$$

Null point on the line joining two charges

• Two like charges $q_1$ and $q_2$ a distance $d$ apart: the field vanishes between them, at distance $x$ from $q_1$ where $\dfrac{q_1}{x^2} = \dfrac{q_2}{(d-x)^2}$. For $+1\,\mu\text{C}$ and $+4\,\mu\text{C}$ placed $30\,\text{cm}$ apart, $\dfrac{1}{x^2} = \dfrac{4}{(30-x)^2}$ gives $30-x = 2x$, so $x = 10\,\text{cm}$ from the $1\,\mu\text{C}$ charge.
• Two unlike charges: the null point lies outside the pair, on the side of the smaller magnitude. For $+8q$ at $x=0$ and $-2q$ at $x=L$, take the point at distance $r$ beyond $-2q$: $\dfrac{8q}{(L+r)^2} = \dfrac{2q}{r^2}$ gives $2r = L+r$, so $r = L$ and the null point is at $x = 2L$.
• Always take the square root of both sides only after checking which region can support a balance.

Field on the axis of a uniformly charged ring — a set 5-mark derivation

A thin ring of radius $r$ carries uniform linear charge density $\lambda$. Find $E$ at a point $P$ on the axis, at distance $x$ from the centre.

Take an element of length $dl$ on the ring, carrying charge $dq = \lambda\,dl$. Its distance from $P$ is $\sqrt{r^2+x^2}$, so the field it produces at $P$ has magnitude

$$dE = \dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{(r^2+x^2)}$$

Resolve $dE$ into a component $dE_x$ along the axis and one perpendicular to it. For every element there is a diametrically opposite element whose perpendicular component is equal and opposite, so by symmetry all perpendicular components cancel round the ring and only the axial components survive:

$$dE_x = dE\cos\theta = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda\,dl}{(r^2+x^2)}\cdot\dfrac{x}{(r^2+x^2)^{1/2}} = \dfrac{\lambda\,x\,dl}{4\pi\varepsilon_0(r^2+x^2)^{3/2}}$$

Integrating round the whole circumference $l = 2\pi r$,

$$E = \int dE\cos\theta = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lambda\,x\,l}{(r^2+x^2)^{3/2}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qx}{(r^2+x^2)^{3/2}}$$

with $q = \lambda l = \lambda(2\pi r)$ the total charge on the ring, directed along the axis away from the ring for $q>0$.

For $x\gg r$, $(r^2+x^2)^{3/2}\approx x^3$, so

$$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qx}{x^3} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}$$

which is exactly the field of a point charge $q$ at distance $x$ — at large distances the ring behaves as a point charge. At the centre, $x = 0$ gives $E = 0$.

Motion of a charge in a uniform field

Since $\vec{F} = q\vec{E}$ is constant, the acceleration $\vec{a} = \dfrac{q\vec{E}}{m}$ is constant and the ordinary kinematic equations apply.

Worked pattern: a particle of charge $2\,\mu\text{C}$ and mass $1.6\,\text{g}$ moving with $\vec{u} = 4\hat{i}\,\text{m s}^{-1}$ enters a field $\vec{E} = 80\hat{i}+60\hat{j}\ \text{N C}^{-1}$ at $t=0$. From $q\vec{E} = m\vec{a}$,

$$\vec{a} = \dfrac{2\times10^{-6}(80\hat{i}+60\hat{j})}{1.6\times10^{-3}} = 0.1\hat{i}+0.075\hat{j}\ \text{m s}^{-2}$$

and $\vec{v} = \vec{u}+\vec{a}t$ at $t = 5\,\text{s}$ gives $\vec{v} = 4.5\hat{i}+0.375\hat{j}\ \text{m s}^{-1}$.

Charged bob in a horizontal field

A small object of charge $q$ and weight $mg$ hangs from a string of length $L$ in a uniform horizontal field $E$, the string settling at angle $\theta$ to the vertical. Resolving the tension: $T\cos\theta = mg$ vertically and $T\sin\theta = qE$ horizontally. Dividing,

$$qE = mg\tan\theta\ \Rightarrow\ q = \left(\dfrac{mg}{E}\right)\tan\theta$$

If the bob is deflected in the direction of $\vec{E}$, the charge is positive.

Standard results

• $E$ of a point charge falls as $1/r^2$; the field of a short dipole falls as $1/r^3$.
• Field due to a charge $q$ at the surface of a sphere it is spread over is $\dfrac{q}{4\pi\varepsilon_0R^2}$.
• A uniformly charged shell behaves outside itself as a point charge at its centre. Combining that with the point-charge law: if a point charge $2q$ gives $E$ at distance $r$, a shell of total charge $q$ gives $\dfrac{kq}{(r/2)^2} = \dfrac{4kq}{r^2} = 2E$ at distance $r/2$, for $r\gg R$.