Suspended Charged Spheres - Detailed Solution & Interactive Quiz

Charged Spheres Equilibrium Problem [GUJCET 2014]

Question: Two spheres having the same radius and mass are suspended by two strings of equal length from the same point in such a way that their surfaces touch each other. When a total charge of $4\times {10}^{-6}\text{C}$ is deposited on them, the charge divides equally so that each sphere gets $2\times {10}^{-6}\text{C}$. They repel each other and, in equilibrium, the angle between their strings becomes ${60}^{\circ }$. Given that the distance from the point of suspension to the center of a sphere is $10\text{cm}$, find the mass of each sphere. (Take $k=9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$ and $g=10{\text{m/s}}^2$.)

(a) $0.3117\text{kg}$

(b) $0.6235\text{kg}$

(c) $0.1559\text{kg}$

(d) $1.2468\text{kg}$

Detailed Step-by-Step Explanation

Step 1: Setup of the Problem
Two identical spheres of mass $m$ are suspended by strings of length $L=10\text{cm}=0.1\text{m}$. In equilibrium, the angle between the strings is ${60}^{\circ }$. This means each string makes an angle of ${30}^{\circ }$ with the vertical.

Step 2: Force Analysis
For each sphere, the forces acting are:
- Weight: $mg$ (downward)
- Tension in the string: $T$ (along the string)
- Electrostatic repulsive force: $F_e$ (horizontal)

Resolving the tension into vertical and horizontal components:

Tension vertical component: $T\cos {30}^{\circ }=mg$   (1)
Tension horizontal component: $T\sin {30}^{\circ }=F_e$  (2)

Dividing (2) by (1):

$\frac{T\sin {30}^{\circ }}{T\cos {30}^{\circ }}=\frac{F_e}{mg}\Rightarrow \tan {30}^{\circ }=\frac{F_e}{mg}$

Since $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$, we have:

$F_e=\frac{mg}{\sqrt{3}}$   (3)

Step 3: Electrostatic Force
Each sphere carries a charge $q=\frac{4\times {10}^{-6}}{2}=2\times {10}^{-6}\text{C}$. The electrostatic repulsive force between two point charges separated by a distance $d$ is given by:

$F_e=\frac{k{q}^2}{d^2}$

Step 4: Distance Between the Centers
Each sphere is displaced horizontally by:

$L\sin {30}^{\circ }=0.1\times 0.5=0.05\text{m}$

Hence, the distance between the centers is:

$d=2\times 0.05=0.1\text{m}$

Step 5: Equating Forces
From (3) and the expression for $F_e$:

$\frac{mg}{\sqrt{3}}=\frac{k{q}^2}{d^2}$

Substitute $d=0.1\text{m}$, $k=9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$, and $g=10{\text{m/s}}^2$:

$\frac{mg}{\sqrt{3}}=\frac{9\times {10}^9\times (2\times {10}^{-6}{)}^2}{(0.1{)}^2}$

Step 6: Solve for $m$
First, calculate $q^2$:
$(2\times {10}^{-6}{)}^2=4\times {10}^{-12}$

And $(0.1{)}^2=0.01$. Thus:

$\frac{mg}{\sqrt{3}}=\frac{9\times {10}^9\times 4\times {10}^{-12}}{0.01}$

Simplify the right-hand side:
$9\times {10}^9\times 4\times {10}^{-12}=36\times {10}^{-3}=0.036$
Dividing by $0.01$:
$\frac{0.036}{0.01}=3.6$

So we have:

$\frac{mg}{\sqrt{3}}=3.6$

Solve for $m$:

$m=\frac{3.6\sqrt{3}}{g}$
Substitute $g=10$:
$m=\frac{3.6\sqrt{3}}{10}$

Now, using $\sqrt{3}\approx 1.732$:

$m\approx \frac{3.6\times 1.732}{10}=\frac{6.2352}{10}=0.62352\text{kg}$

Final Answer: The mass of each sphere is approximately $0.6235\text{kg}$, which corresponds to option (b).