Question

Two identical balls each of mass $\frac{40}{3}$ g carry equal and opposite charge. They are suspended from a horizontal plate by silk threads each of length 1 m with a separation of 1.5 m between points of suspension. At equilibrium, if the distance between the balls is 30 cm then magnitude of charge on each ball is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

• A. $10^{-5} \text{ C}$
• B. $10^{-6} \text{ C}$
• C. $10^{-3} \text{ C}$
• D. $2\times 10^{-6} \text{ C}$

Hint:

Geometry is key: find the horizontal offset per string $x = \frac{1.5 - 0.3}{2} = 0.6\text{ m}$. With $L=1\text{m}$, this forms a classic 3-4-5 right triangle ($0.6, 0.8, 1$), so $\tan\theta = 3/4$.

Answer 1

The Correct Option is B

Step 1: Balance of forces.
Each charged ball hangs in balance, so $\tan\theta = \dfrac{F_{e}}{mg}$, where $F_{e}$ is the electric force.

Step 2: Find the sideways shift.
Suspension points are $1.5$ m apart, balls are $0.3$ m apart. Each thread shifts sideways by $x = \dfrac{1.5 - 0.3}{2} = 0.6$ m.

Step 3: Find the height and $\tan\theta$.
Thread length $1$ m, so height $h = \sqrt{1 - 0.36} = 0.8$ m. Then $\tan\theta = \dfrac{0.6}{0.8} = \dfrac{3}{4}$.

Step 4: Find the electric force.
With $m = \dfrac{40}{3}\times 10^{-3}$ kg and $g=10$: \[ F_{e} = mg\tan\theta = \left(\tfrac{40}{3}\times 10^{-3}\right)(10)\left(\tfrac{3}{4}\right) = 0.1 \text{ N} \]
Step 5: Use Coulomb's law.
\[ \frac{9\times 10^{9}\,q^{2}}{(0.3)^{2}} = 0.1 \]so $10^{11}q^{2} = 0.1$.

Step 6: Solve for $q$.
\[ q^{2} = 10^{-12} \quad\Rightarrow\quad q = 10^{-6} \text{ C} \]This is option 2.
\[ \boxed{q = 10^{-6} \text{ C}} \]