Define electric field strength and give its unit of measurement - Leaving Cert Physics - Question d - 2010

In the diagram, the electric field at point P is directed away from the positive charge (5 µC) and towards the negative charge (−2 µC). The direction of the electric field vector, denoted as ( \vec{E} ), is indicated as follows:

(Assuming the diagram is appropriately copied into the answerbook.)

To calculate the electric field strength at point P, we need to consider the contributions from both charges.

Using the formula for electric field strength:

$E = \frac{q}{4\pi \epsilon_0 d^2}$

The total electric field at point P, due to both charges:

1. From Charge -2 µC:

   • Distance to P from -2 µC = 10 cm = 0.1 m

   $E_{-2\mu C} = \frac{2 \times 10^{-6}}{4\pi \times 8.9 \times 10^{-12} \times (0.1)^2}$

2. From Charge 5 µC:

   • Distance to P from 5 µC = 15 cm = 0.15 m

   $E_{5\mu C} = \frac{5 \times 10^{-6}}{4\pi \times 8.9 \times 10^{-12} \times (0.15)^2}$

After calculating both fields:

Total electric field strength at P:

$E_{total} = E_{-2\mu C} + E_{5\mu C}$

The result is approximately:

$E_{total} = 3.77 \times 10^4 \text{ N C}^{-1}$

(Directions indicated).

Point discharge will occur under conditions of:

1. A large electric field strength at a point.
2. A high charge density at that point.

When these conditions are met, the electric field is strong enough to cause ionization in the surrounding medium, leading to a discharge.