Three small identical metal spheres, P, S and T, on insulated stands, are initially neutral - NSC Physical Sciences - Question 7 - 2018 - Paper 1

When the charged sphere S (with charge -3 x 10⁹ C) and the charged sphere T (with charge +2 x 10⁹ C) are separated, the electric field lines would radiate out from sphere T and converge inward toward sphere S.

• The positive sphere (T) has electric field lines that point outward, indicating a field directed away from it.
• The negative sphere (S) has electric field lines directed toward it, indicating that it attracts positive charges towards itself.

In a diagram, the electric field lines should not cross, and they should touch the spheres.

Coulomb's law states that the force 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 magnitude of the electrostatic force between the charges,
• $k$ is Coulomb's constant, approximately equal to $8.99 x 10^9 N m^2/C^2$, and
• $r$ is the distance between the charges.

To calculate the net electrostatic force acting on sphere P, we need to find the individual forces due to spheres S and T and then sum them vectorially.

• The force due to sphere S on P:

$F_{SP} = k \frac{|q_S q_P|}{d_{SP}^2} = (8.99 x 10^9) \frac{(3 x 10^{-9})(3 x 10^{-9})}{(0.1)^2}$

• The force due to sphere T on P:

$F_{TP} = k \frac{|q_T q_P|}{d_{TP}^2} = (8.99 x 10^9) \frac{(2 x 10^{-9})(3 x 10^{-9})}{(0.3)^2}$

Finding the net force:

$F_{net} = F_{SP} + F_{TP}$

To calculate the net electric field at the origin due to charges S and T, we must find the electric field due to each charge. The electric field due to a point charge is given by:

$E = k \frac{|q|}{r^2}$

1. Electric field due to charge S at the origin:

$E_S = k \frac{|q_S|}{d_{SP}^2} = (8.99 x 10^9) \frac{(3 x 10^{-9})}{(0.1)^2}$

2. Electric field due to charge T at the origin:

$E_T = k \frac{|q_T|}{d_{TP}^2} = (8.99 x 10^9) \frac{(2 x 10^{-9})}{(0.3)^2}$

The net electric field is the vector sum of these two fields.

The sphere that experienced a very small increase in mass is sphere T, because as it gains positive charge, it attracts electrons from its surroundings, leading to an increase in its mass.

To calculate the increase in mass due to the charge gained by sphere T, we can use the formula:

$m_{gained} = \frac{q}{e}$ where $q$ is the charge gained and $e$ is the elementary charge (approximately $1.60 x 10^{-19} C$).

For sphere T, the charge gained is $2 x 10^{-9} C$:

$m_{gained} = \frac{2 x 10^{-9} C}{1.60 x 10^{-19} C} \times 9.11 x 10^{-31} kg = 1.125 x 10^{-10} kg.$