Radial field strength follows an inverse square law

Electric Fields & Potential — AQA A-Level Physics

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

$E$: electric field strengthThe force per unit positive chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. (N C^-1)
$Q$: point chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). producing the field (C)
$r$: distance from the centre of the chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). (m)
$varepsilon_0$: permittivity of free space (F m^-1)

Derived by substituting Coulomb's lawThe electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. into $E = F/Q$: the test charge q cancels.
E is not constant in a radial field. It decreases with the square of the distance.
For a positive charge, E is positive (points outward). For a negative charge, E is negative (points inward).
The E-r graph is a 1/$r^{2}$ curve that starts high near the charge and drops off rapidly.

Combining electric fields

Electric field strengthThe force per unit positive charge at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. is a vector. Multiple fields combine by vector addition.
On the same line: add if in the same direction, subtract if in opposite directions.
At right angles: use Pythagoras, $E_{\text{resultant}} = \sqrt{E1^2 + E2^2}$.

Worked Example

A metal sphere of diameter 15 cm has an electric field strength of $1.5 \times 10^{5}$ V m^-1 at its surface. Determine the total surface charge.

Show Solution

List known values

$E = 1.5 \times 10^{5} \text{ V m}^{-1}$
Radius $r = 7.5 \text{ cm} = 7.5 \times 10^{-2} \text{ m}$
$\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$

Rearrange E = Q/(4piepsilon0 $r^{2}$) for Q

$$Q = 4\pi\varepsilon_0 E r^2$$

Substitute and calculate

$$Q = (4\pi \times 8.85 \times 10^{-12}) \times (1.5 \times 10^{5}) \times (7.5 \times 10^{-2})^2$$ $$= 9.4 \times 10^{-8} \text{ C} = 94 \text{ nC}$$

Answer

$Q = 94$ nC