A small neutral sphere acquires a charge of -1,95 × 10⁻⁶ C - NSC Physical Sciences - Question 7 - 2021 - Paper 1

To calculate the number of electrons, we use the charge of a single electron, which is approximately -1,6 × 10⁻¹⁹ C. The number of electrons (n) can be calculated using the formula:

$n = \frac{Q}{e} = \frac{-1.95 \times 10^{-6} C}{-1.6 \times 10^{-19} C/e} \approx 1.22 \times 10^{13} \, \text{electrons}$

The electric field at a point in space is defined as the force per unit charge experienced by a small positive test charge placed at that point. It is represented by the formula:

$E = \frac{F}{q}$

Where, $E$ is the electric field, $F$ is the force acting on the test charge, and $q$ is the magnitude of the charge.

The electric field (E) due to a point charge is calculated using the formula:

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

Where:

• $k \approx 8.99 \times 10^9 N m^2/C^2$ (Coulomb's constant)
• $Q = -1.95 \times 10^{-6} C$
• $r = 0.5 m$

Substituting the values:

$E = \frac{(8.99 \times 10^9) \cdot (1.95 \times 10^{-6})}{(0.5)^2} \approx 7.18 \times 10^4 N/C$.

Using Coulomb's Law, the net electrostatic force (${F}_{net}$) on charge $q_1$ can be expressed as:

$F_{net} = F_{21} + F_{12}$

Where:

• ${F}_{12} = k \frac{q_1 q_2}{r^2}$ (force between q₁ and q₂)
• ${F}_{21} = k \frac{q_2 q_1}{r^2}$ (force between q₂ and q₁)

Given that:

• $F_{net} = -1.38 N$ (west direction implies negative)
• The distance between $q_1$ and $q_2$ is $0.03 m$.

Through proper substitution and rearrangement, we solve for $q_2$, leading to:

$q_2 = \frac{F_{net} r^2}{k |q_1|}$

The value for $q_2$ can be calculated, eventually yielding:

$q_2 = 1.11 \times 10^{-7} C$.