18.4 Electric field of a point charge

Electric Field Strength of a Radial Field

• Unlike the uniform field between plates ($\mathrm{E}=\frac{\mathrm{\Delta V}}{\mathrm{\Delta d}}$), the field around a point charge (or spherical conductor) is radial and non-uniform.
• Its strength decreases rapidly with distance (inverse square law).

Formula: Radial Field Strength

The electric field strength $E$ at a distance $r$ from a point charge $Q$ is:

$E=\frac{Q}{4\pi {ϵ}_0{r}^2}$

• $E$: Electric Field Strength (N C${}^{-1}$ or V m${}^{-1}$)
• $Q$: Magnitude of the source charge (Coulombs, C)
• $r$: Distance from the centre of the charge (Metres, m)
• ${ϵ}_0$: Permittivity of Free Space ($8.85\times 10^{-12}$ F m${}^{-1}$)

Derivation from Coulomb’s Law

• Recall the definition of electric field strength: Force per unit positive charge:

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

• Recall Coulomb’s Law for the force between source charge, $Q$, and test charge, $q$:

$$\mathrm{F}=\frac{Qq}{4\pi {ϵ}_{\circ }{r}^2}$$

• Substitute $\mathrm{F}$:

$$\mathrm{E}=\frac{Qq}{4\pi {ϵ}_{\circ }{r}^2}\times \frac{1}{q}=\frac{Q}{4\pi {ϵ}_{\circ }{r}^2}$$

Vector Nature

• $E$ is a vector.
• If $Q$ is Positive: $E$ points radially outwards (Repels a $+ve$ test charge).
• If $Q$ is Negative: $E$ points radially inwards (Attracts a $+ve$ test charge).

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Graphical Representation (Inverse Square Law)

The graph of $\mathrm{E}$ against $r$ is an Inverse Square Relationship ($\mathrm{E}\propto \frac{1}{r^2}$).

Electric field strength is zero inside a charged sphere and decreases with distance outside the sphere according to an inverse square law

Sketching the Graph

Scenario: A positively charged conducting sphere of radius $R$.

1. Region $0\to R$: $E=0$ (Inside a conductor).
2. At $r=R$: $E=E_{max}=\frac{Q}{4\pi {ϵ}_0{R}^2}$ (Maximum value at the surface).
3. Region $r>R$: Curve decays rapidly.
   • At $2R$, field is $E_{max}/4$.
   • At $3R$, field is $E_{max}/9$.
   • The curve is asymptotic to the r-axis (never touches 0).

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Superposition of Electric Fields

For multiple charges, the total field strength at a point is the vector sum of the individual field strengths:

$${\mathrm{E}}_{\text{total}}={\mathrm{E}}_1+{\mathrm{E}}_2+\dots$$

For charges along the same line, the resultant field is the vector addition of the field due to both charges at a particular point

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Isolation of Variables (Ratio Questions)

Formula for ratio of new/old charge and distance

Formula

$\frac{{\mathrm{E}}_{new}}{{\mathrm{E}}_{old}}=\frac{{\mathrm{Q}}_{new}}{{\mathrm{Q}}_{old}}\times {\left(\frac{r_{old}}{r_{new}}\right)}^2$

Derivation

$$\begin{array}{rl}\mathrm{E}& \propto \frac{\mathrm{Q}}{r^2}\\ \mathrm{E}& =k\frac{\mathrm{Q}}{r^2}\\ k& =\frac{\mathrm{E}{r}^2}{\mathrm{Q}}\\ k_{new}& =k_{old}\\ \frac{{\mathrm{E}}_{new}{r}_{new}^2}{{\mathrm{Q}}_{new}}& =\frac{{\mathrm{E}}_{old}{r}_{old}^2}{{\mathrm{Q}}_{old}}\\ \frac{{\mathrm{E}}_{new}}{{\mathrm{E}}_{old}}& =\frac{{\mathrm{Q}}_{new}}{{\mathrm{Q}}_{old}}\times {\left(\frac{r_{old}}{r_{new}}\right)}^2\end{array}$$

Example

At a distance of $x$ from a charge $Q$, the electric field strength is $\mathrm{E}$. What is the field strength at a distance $3x$ from a charge $2Q$?

Method:

$\frac{E_{new}}{E_{old}}=\frac{Q_{new}}{Q_{old}}\times {\left(\frac{r_{old}}{r_{new}}\right)}^2$

$\frac{E_{new}}{E}=\frac{2Q}{Q}\times {\left(\frac{x}{3x}\right)}^2=2\times \frac{1}{9}=\frac{2}{9}$

$\therefore E_{new}=\frac{2}{9}E$

Common Mistakes

1. Square the Distance: Remember to square the distance.
2. Radius vs Distance: If asked for the field at a height $h$ above a sphere of radius $R$, the distance is $r=R+h$.