In Chapter 21, we saw that the electric field strength at a point is defined as the force per unit charge exerted on a positive charge placed at that point, $E = \frac{F}{Q}$.
So, to find the field strength near a point charge ${{Q_1}}$ (or outside a uniformly charged sphere), we have to imagine a small positive test charge ${{Q_2}}$ placed in the field, and determine the force per unit charge on it.
We can then use the definition to determine the electric field strength for a point (or spherical) charge.
The force between the two point charges is given by:

$F = \frac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$

The electric field strength E due to the charge ${{Q_1}}$ at a distance of r from its centre is thus:

$\begin{array}{l}
E = \frac{{force}}{{test\,ch\arg e}}\\
 = \frac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{r^2}{Q_2}}}\\
 \equiv \frac{Q}{{4\pi {\varepsilon _0}{r^2}}}
\end{array}$

where E is the electric field strength due to a point charge Q, and r is the distance from the point.
The field strength E is not a constant; it decreases as the distance r increases. The field strength obeys an inverse square law with distance–just like the gravitational field strength for a point mass. The field strength will decrease by a factor of four when the distance from the centre is doubled.
Note also that, since force is a vector quantity, it follows that electric field strength is also a vector. We need to give its direction as well as its magnitude in order to specify it completely. Worked example 1 shows how to use the equation for field strength near a charged sphere.

WORKED EXAMPLE

1) A metal sphere of diameter $12 cm$ is positively charged. The electric field strength at the surface of the sphere is $4.0 \times {10^5}\,V\,{m^{ - 1}}$. Draw the electric field pattern for the sphere and determine the total surface charge.

 

Step 1: Draw the electric field pattern (Figure 22.5). The electric field lines must be normal to the surface and radial.
Step 2: Write down the quantities given:
electric field strength $E = 4.0 \times {10^5}\,V\,{m^{ - 1}}$
$\begin{array}{l}
rudius\,r = \frac{{0.12}}{2}\\
 = 0.06\,m
\end{array}$
Step 3: Use the equation for the electric field strength to determine the surface charge:
$\begin{array}{l}
E = \frac{Q}{{4\pi {\varepsilon _0}{r^2}}}\\
Q = 4\pi {\varepsilon _0}{r^2} \times E\\
 = 4\pi  \times 8.85 \times {10^{ - 12}} \times {(0.06)^2} \times 4.0 \times {10^5}\\
 = 1.6 \times {10^{ - 7}}\,C\\
 \equiv 0.16\,\mu C
\end{array}$