Define electric field strength - Leaving Cert Physics - Question (b) - 2016

The electric field pattern between two oppositely charged parallel plates consists of equally spaced parallel lines that extend from the positive plate to the negative plate. This indicates a uniform electric field.

When the drop of oil is stationary, there are two forces acting on it:

1. Force of Weight (W): This force acts downwards due to gravity and can be expressed as: $W = mg$ where m is the mass and g is the acceleration due to gravity.

2. Equal Upward Force (F_e): This force acts upwards due to the electric field, balancing the downward weight of the drop. Hence, these two forces are equal in magnitude but opposite in direction.

Given:

• Electric field strength, $E = 3.6 \times 10^4 \text{ V m}^{-1}$
• Mass of the drop, $m = 2.4 \times 10^{-15} \text{ kg}$

First, we calculate the force on the drop due to gravity:

$F = mg = (2.4 \times 10^{-15} ext{ kg})(9.8 ext{ m s}^{-2}) = 2.352 \times 10^{-14} ext{ N}$

Now, using the relationship $F = Eq$, we can rearrange it to find the charge (q):

$q = \frac{F}{E} = \frac{2.352 \times 10^{-14} ext{ N}}{3.6 \times 10^4 ext{ V m}^{-1}} = 6.53 \times 10^{-19} ext{ C}$

To find the number of excess electrons (n) on the drop, we use the charge of a single electron, which is approximately $1.6 \times 10^{-19} ext{ C}$:

$n = \frac{q}{e} = \frac{6.53 \times 10^{-19} ext{ C}}{1.6 \times 10^{-19} ext{ C}} \approx 4.1$

Rounding to the nearest integer, there are approximately 4 excess electrons on the drop.