Define electric field strength - Leaving Cert Physics - Question 10 - 2005

Coulomb's law states that the force (F) 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 (r) between them. It is given by the formula:

$F = k \frac{|Q_1 Q_2|}{r^2}$

Where:

• F = force between the charges (N)
• Q1, Q2 = magnitudes of the charges (C)
• r = distance between the charges (m)
• k = Coulomb's constant (approximately $8.99 \times 10^9 \, N \, m^2/C^2$)

Coulomb's law is considered an inverse square law because the force between two electric charges decreases with the square of the distance between them. Specifically, when the distance is doubled, the force is reduced by a factor of four, illustrating that force is proportional to ( \frac{1}{r^2} ).

1. Magnitude: The gravitational force is significantly weaker than the electrostatic force; for example, the electrostatic force between two electrons is much stronger compared to the gravitational force between them.

2. Nature of Force: The gravitational force is always attractive, while the electrostatic force can be either attractive or repulsive, depending on the charges of the interacting particles.

To demonstrate electric field lines, one can use a high-voltage power supply connected to two metal plates (electrodes).

1. Arrange a semi-fluid and oil container to allow for the deposition of charged particles.
2. Connect the high voltage to the plates (placed in the container).
3. Observe that the semolina aligns along the electric field lines, visually depicting the electric field pattern.

Using Coulomb's law, the electric field strength (E) at point B can be calculated as follows:

$E = \frac{F}{Q} = \frac{Q}{4\pi \epsilon_0 r^2}$

Substituting the known values:

• Charge on the electron, Q = $1.6 \times 10^{-19}\, C$
• Permittivity of free space, $\epsilon_0 = 8.9 \times 10^{-12}\, F/m$
• Distance, r = 10 mm = 0.01 m

So, we have:

$E = \frac{(1.6 \times 10^{-19})}{4 \pi (8.9 \times 10^{-12})(0.01)^2}$

Calculating this gives:

$E \approx 1.44 \times 10^6 \, N/C$

The direction of the electric field strength at point B is towards the electron since electric field lines point from positive to negative charges.

To find the electrostatic force (F) exerted on a charge placed in an electric field, we can use the formula:

$F = E \times Q$

Where:

• E is the electric field strength calculated previously ($1.44 \times 10^6 \, N/C$)
• Q is the charge placed at B, which is 5 μC = $5 \times 10^{-6} C$

Thus, substituting the known values:

$F = (1.44 \times 10^6) \times (5 \times 10^{-6})$

Calculating this gives:

$F \approx 7.2 \, N$