7.1 Draw the electric field pattern due to the +4 nC charge - NSC Physical Sciences - Question 7 - 2024 - Paper 1

The electric field E due to a point charge is given by the formula: $E = k \frac{Q}{r^2}$

Where:

• $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$ (Coulomb's constant)
• $Q = 4 \text{ nC} = 4 \times 10^{-9} \text{ C}$
• $r = 0.025 \text{ m}$

Substituting these values into the formula: $E = 9 \times 10^9 \times \frac{4 \times 10^{-9}}{(0.025)^2}$

Calculating the denominator: $(0.025)^2 = 0.000625$

Thus, we have: $E = 9 \times 10^9 \times \frac{4 \times 10^{-9}}{0.000625} = 5.76 \times 10^4 \text{ N C}^{-1}$

Therefore, the electric field at point X is approximately $57,600 \text{ N C}^{-1}$.

Coulomb's Law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

First, determine the weight of each ball: $W = mg$ Where:

• $m = 0.012 \text{ kg}$
• $g = 9.8 \text{ m/s}^2$ (acceleration due to gravity)

$W = 0.012 \times 9.8 = 0.1176 \text{ N}$

Next, the tension in the string $T$ can be expressed in terms of angles, where the angle with respect to the vertical is 9°: $T = \frac{W}{\cos(9)}$

The vertical component of the tension ($T_y$) should equal the weight: $T_y = T \cdot \cos(9) = 0.1176 \text{ N}$

The electric force $F_E$ acting on ball B when charges are equalized: $F_E = k \frac{Q_A Q_B}{d^2}$ Where $d=0.1 ext{ m}$ and $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

To solve for $Q_B$, set the equation for the electric force equal to the tension force when the balls are at rest: $F_E = T_y$ $k \frac{Q^2}{(0.1)^2} = 0.1176 \text{ N}$

Solving gives: $Q^2 = \frac{0.1176 \cdot (0.1)^2}{9 \times 10^9}$

Calculating further yields: $Q^2 = 1.44 \times 10^{-7}$ Therefore, $Q = \sqrt{1.44 \times 10^{-7}} \approx 1.2 \times 10^{-4} \text{ C}$ Thus, the charge given to ball B is approximately $1.2 \times 10^{-4} \text{ C}$.