Question

$\ell = 30m$

$\underline{m}$ ?

Answer 1

$m \approx 8.55\times10^7\,\mathrm{kg}$.

Answer 2

We have two identical spheres (mass = m) hanging by strings of length ℓ = 30 m. They carry charges 3 C (left) and 2 C (right) and are in electrostatic equilibrium. The strings make an angle of 15° with the vertical (since the total angle between them is 30°).

For each sphere, resolve the forces:

• Weight: $mg$ downward.
• Tension: $T$ along the string.
• Electrostatic repulsion: $F_e$ horizontally.

In equilibrium,

$$T\cos15^\circ = mg \quad \text{and} \quad T\sin15^\circ = F_e.$$

Dividing,

$$\tan15^\circ = \frac{F_e}{mg} \quad\Longrightarrow\quad m = \frac{F_e}{g\,\tan15^\circ}.$$

The horizontal repulsive force between the spheres is given by Coulomb’s law:

$$F_e = k\,\frac{(3)(2)}{d^2} = \frac{6k}{d^2},$$

where $k = 9\times10^9 \,\mathrm{N\,m^2/C^2}$ and the separation $d$ is determined by the horizontal displacement of each sphere:

$$d = 2\ell\sin15^\circ.$$

Thus,

$$d^2 = 4\ell^2\sin^2 15^\circ.$$

So,

$$F_e = \frac{6k}{4\ell^2\sin^2 15^\circ} = \frac{3k}{2\ell^2\sin^2 15^\circ}.$$

Substitute $F_e$ into the m-equation:

$$m = \frac{3k}{2\ell^2\sin^2 15^\circ\,g\,\tan15^\circ}.$$

Inserting numerical values:

• ℓ = 30 m,
• $k = 9\times10^9 \,\mathrm{N\,m^2/C^2}$,
• $g = 9.8\,\mathrm{m/s^2}$,
• $\sin15^\circ \approx 0.258819$ so that $\sin^2 15^\circ \approx (0.258819)^2 \approx 0.066987$,
• $\tan15^\circ \approx 0.267949$.

Now,

$$m = \frac{3\times9\times10^9}{2\times(30)^2\times0.066987\times9.8\times0.267949}.$$

Calculate step‐by‐step:

1. $(30)^2 = 900$.
2. Denominator factor: $2 \times 900 = 1800$.
3. Then, $1800 \times 0.066987 \approx 120.376$.
4. Next, $120.376 \times 9.8 \approx 1179.725$.
5. And then, $1179.725 \times 0.267949 \approx 315.75$.

The numerator is:

$$3\times9\times10^9 = 27\times10^9.$$

Thus,

$$m \approx \frac{27\times10^9}{315.75} \approx 85.5\times10^6\,\mathrm{kg} = 8.55\times10^7\,\mathrm{kg}.$$