Electric Field

No. The tangent to a line of force at any point gives the direction of the electric field at that point. If two lines crossed, there would be two tangents, that is two different directions of the field, at the point of intersection. But the net field at a point has one definite direction (a test charge can only move along one path), so lines of force can never intersect.

Electric field intensity: the electric force experienced per unit positive test charge placed at a point, $E=\dfrac{F}{q_0}$ (unit N C$^{-1}$ or V m$^{-1}$).

Field due to an infinite plane sheet of charge: Let the sheet have uniform surface charge density $\sigma$. By symmetry the field is perpendicular to the plane and points away from it (if $\sigma>0$), with the same magnitude $E$ at equal distances on either side.

Use Gauss's law with a cylindrical Gaussian surface of cross-section $A$, with its axis perpendicular to the sheet and its ends equidistant on either side. The field is parallel to the curved surface, so the flux there is zero. Flux through each end is $EA$, so the total flux is $2EA$. The enclosed charge is $\sigma A$.

$2EA=\frac{\sigma A}{\varepsilon_0}\ \Rightarrow\ E=\frac{\sigma}{2\varepsilon_0}$

The field is independent of the distance from the sheet. For a charged conducting plate, the charge spreads over both faces; the field just outside the surface is $E=\dfrac{\sigma}{\varepsilon_0}$, where $\sigma$ is the charge density on that face.

(a) electric field only

A stationary charge produces only an electric field. A magnetic field needs a moving charge (current), and radiated fields need an accelerating charge.

(a) Field lines of two identical negative charges:

• The lines run towards each charge, so the arrows point inward and the lines end on the charges. They come in from far away.
• Close to each charge the lines are radial, as for a single negative charge.
• The two charges repel, so lines from the two sides never meet or cross. They bend away from each other, leaving a gap between the charges.
• Midway between the charges there is a neutral point where the resultant field is zero.
• Far away the pattern looks like that of a single charge of $-2Q$.

(b) Take X at distance $r$ from each charge $-Q$ on the perpendicular bisector (the equilateral arrangement of Fig. 6.1). Each charge produces a field of magnitude

$E_1=E_2=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}$

directed towards that charge. Components parallel to the line joining the charges cancel. Components along the bisector add. Each field makes $30^\circ$ with the bisector:

$E=2E_1\cos30^\circ=\frac{2}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\cdot\frac{\sqrt3}{2}=\frac{\sqrt3\,Q}{4\pi\varepsilon_0 r^{2}}$

The resultant is directed along the bisector, towards the midpoint of the line joining the two charges.

Field pattern of two identical negative point charges:

• The lines run towards each charge, so the arrows point inward and the lines end on the charges. They come in from far away.
• Close to each charge the lines are radial, as for a single negative charge.
• The two charges repel, so lines from the two sides never meet or cross. They bend away from each other, leaving a gap between the charges.
• Midway between the charges there is a neutral point where the resultant field is zero.
• Far away the pattern looks like that of a single charge of $-2Q$.

This is the pattern for two positive charges with every arrow reversed.

Take X at distance $r$ from each charge $-Q$ on the perpendicular bisector (the equilateral arrangement of Fig. 6.1). Each charge produces a field of magnitude

$E_1=E_2=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}$

directed towards that charge. Components parallel to the line joining the charges cancel, and components along the bisector add. Each field makes $30^\circ$ with the bisector:

$E=2E_1\cos30^\circ=\frac{\sqrt3\,Q}{4\pi\varepsilon_0 r^{2}}$

The resultant is directed along the bisector, towards the midpoint of the line joining the two charges.

Gauss's theorem. The total electric flux through any closed surface equals $\dfrac{1}{\varepsilon_0}$ times the net charge enclosed by it:

$\oint \vec E\cdot d\vec A=\frac{q}{\varepsilon_0}$

The surface is called a Gaussian surface. It is chosen so that $E$ is constant and perpendicular to it, which makes the integral easy.

Field of a charged sphere of radius $R$ and total charge $Q$ (spherical symmetry means $E$ is radial and has the same magnitude on any concentric sphere).

(i) Outside ($r>R$). Take a concentric Gaussian sphere of radius $r$. It encloses charge $Q$:

$E\,(4\pi r^2)=\frac{Q}{\varepsilon_0}\;\Rightarrow\;E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$

The sphere acts as if all its charge were at the centre.

(ii) Inside ($r<R$).

• Conducting (or hollow) sphere: all charge lies on the surface, so a Gaussian sphere of radius $r<R$ encloses no charge and $E=0$.
• Uniformly charged solid (non-conducting) sphere: the enclosed charge is $Q\dfrac{r^3}{R^3}$, so $E\,(4\pi r^2)=\frac{Q r^3}{\varepsilon_0R^3}\;\Rightarrow\;E=\frac{1}{4\pi\varepsilon_0}\frac{Qr}{R^3}$

The field is zero inside a conductor, and increases linearly with $r$ inside a uniformly charged solid sphere.