Two small spheres, X and Y, carrying charges of +6 x 10^-6 C and +8 x 10^-6 C respectively, are placed 0.20 m apart in air - NSC Physical Sciences - Question 7 - 2017 - Paper 1

Using Coulomb's law, the force experienced by charged sphere X due to sphere Y can be calculated as follows:

$F_{XY} = k \frac{|Q_X Q_Y|}{r^2}$

Where:

• $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$
• $Q_X = +6 \times 10^{-6} \, \text{C}$
• $Q_Y = +8 \times 10^{-6} \, \text{C}$
• $r = 0.20 \, \text{m}$

Substituting the given values:

$F_{XY} = (9 \times 10^9) \frac{(6 \times 10^{-6})(8 \times 10^{-6})}{(0.20)^2} = 10.8 \, \text{N}$

The vector diagram would show:

• The force $F_{XY}$ acting upwards on sphere Y due to sphere X.
• The force $F_{YZ}$ acting downwards on sphere Y due to sphere Z.
• The net force $F_{net}$ on sphere Y being the vector sum of $F_{XY}$ and $F_{YZ}$.

Given that the net electrostatic force on sphere Y is $15.20 \, \text{N}$ and the force exerted by sphere X is $10.8 \, \text{N}$, we can calculate the force due to sphere Z:

$F_{YZ} = F_{net} - F_{XY} = 15.20 \, \text{N} - 10.8 \, \text{N} = 4.40 \, \text{N}$

Using Coulomb’s law for the force between spheres Y and Z:

$F_{YZ} = k \frac{|Q_Y Q_Z|}{r^2}$

Where:

• $Q_Y = +8 \times 10^{-6} \, \text{C}$
• $r = 0.30 \, \text{m}$

Rearranging gives: $Q_Z = \frac{F_{YZ} \cdot r^2}{k \cdot |Q_Y|}$

Substituting values:

$Q_Z = \frac{4.40 \cdot (0.30)^2}{9 \times 10^9 \cdot 8 \times 10^{-6}} = -1.34 \times 10^{-5} \, \text{C}$