8.3 Electric field: concept of a field revisited  (Page 2/5)

Calculating the electric field of a point charge

Calculate the strength and direction of the electric field $E$ due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

Strategy

We can find the electric field created by a point charge by using the equation $E=\text{kQ}/r^2$ .

Solution

Here $Q=2\text{.}\text{00}\times {\text{10}}^{-9}$ C and $r=5\text{.}\text{00}\times {\text{10}}^{-3}$ m. Entering those values into the above equation gives

Discussion

This electric field strength is the same at any point 5.00 mm away from the charge $Q$ that creates the field. It is positive, meaning that it has a direction pointing away from the charge $Q$ .

Calculating the force exerted on a point charge by an electric field

What force does the electric field found in the previous example exert on a point charge of $\mathrm{–0.250}\mu \text{C}$ ?

Strategy

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field $\mathbf{\text{E}}=\mathbf{\text{F}}/q$ rearranged to $\mathbf{\text{F}}=q\mathbf{\text{E}}$ .

Solution

The magnitude of the force on a charge $q=-0\text{.}\text{250}\text{μC}$ exerted by a field of strength $E=7\text{.}\text{20}\times {\text{10}}^5$ N/C is thus,

Because $q$ is negative, the force is directed opposite to the direction of the field.

Discussion

The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.