A small sphere, Y, carrying an unknown charge is suspended at the end of a light inextensible string which is attached to a fixed point - NSC Physical Sciences - Question 7 - 2019 - Paper 1

To find the charge on sphere Y, we can use Coulomb's Law:

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

Where:

• $F_E = 3,05 \, N$ (the electrostatic force)
• $k = 9 \times 10^9 \, N m^2/C^2$ (Coulomb's constant)
• $q_1 = 6 \times 10^6 \, C$ (charge on sphere X)
• $r = 0.2 \, m$ (distance between the spheres)

Rearranging the formula to solve for $q_2$ (which represents the charge on sphere Y):

$q_2 = \frac{F_E \cdot r^2}{k \cdot q_1}$

Substituting the known values:

$q_2 = \frac{3.05 \, N \cdot (0.2 \, m)^2}{(9 \times 10^9) \cdot (6 \times 10^6)}$

Calculating, we find:
$q_2 = 2.26 \times 10^{-6} \, C$

A free-body diagram for sphere Y should include:

• The tension in the string (T) acting upwards at an angle of 10° to the vertical.
• The weight of the sphere (W = mg) acting downwards.
• The electrostatic force (F_E) acting horizontally towards sphere X.

Label these forces accordingly in the diagram.

To find the tension, we can use the components of forces:

• In the vertical direction: $T \cos(10°) = W$

• In the horizontal direction: $F_E = T \sin(10°)$

Substituting the weight (assumed to be small compared to electrostatic forces) and solving for T:

1. First calculate weight, $W = m imes g$ (assume m is small).
2. Then use the equations above to find T: $T = \frac{F_E}{\sin(10°)}$ Substituting values from earlier: $T = \frac{3.05}{\sin(10°)}$
   Calculating gives us the value of T.