Define the unit of charge, the coulomb - Leaving Cert Physics - Question c - 2013

Coulomb's law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:

$F = k \frac{|q_1 q_2|}{r^2}$

where $F$ is the force between the charges, $k$ is Coulomb's constant ($8.99 \times 10^9 \frac{N m^2}{C^2}$), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges.

To calculate the force of repulsion, we will use Coulomb's law mentioned above.

Given:

• Charge of each sphere, $q_1 = q_2 = +3 \ \mu C = 3 \times 10^{-6} C$
• Distance, $r = 8 \ cm = 0.08 \ m$

Substituting these values into the formula:

$F = k \frac{|q_1 q_2|}{r^2}$

Calculating:

$F = 8.99 \times 10^9 \frac{N m^2}{C^2} \cdot \frac{|(3 \times 10^{-6})(3 \times 10^{-6})|}{(0.08)^2}$

Calculating the numerator:

$= 8.99 \times 10^9 \cdot \frac{9 \times 10^{-12}}{0.0064}$

$= 8.99 \times 10^9 \cdot 1.40625 \times 10^{-9}$

$F \approx 12.64 \ N$

In the diagram, the electric field lines emanate from the positive charges. The region where the electric field strength is zero will be found along the line connecting the two charges. This point is exactly halfway between the two charges, where the fields from each charge cancel each other out. To illustrate this:

• Draw field lines from each sphere pointing away from the charge.
• Mark the point halfway between the two charges as the location where the electric field is zero.