(II) A large electroscope is made with “leaves” that are 78-cm-long wires with tiny 21-g spheres at the ends. When charged, nearly all the charge resides on the spheres. If the wires each make a 26° angle with the vertical (Fig. 16–55), what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

The force between two point charges relies on the magnitude of both the charges and the separation between them.

The expression for the force between two point charges is given as:

\(F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant, \({Q_1},\;{Q_2}\) are the charges and r is the separation between them.

In an equilibrium state, the net force on each sphere must be zero.

The mass of each sphere is,\(m = 21\;{\rm{g}} = 0.021\;{\rm{kg}}\).

The length of the wires is,\(l = 78\;{\rm{cm}} = 0.78\;{\rm{m}}\).

The angle with the vertical is, \(\theta = 26^\circ \).

As both the spheres are identical, the charge will distribute equally on both the spheres.

From the figure you get,

\(\begin{aligned}{c}\frac{d}{l} = \sin \theta \\d = l\sin \theta \end{aligned}\)

The distance between the charges is,

\(\begin{aligned}{c}r = 2d\\ = 2l\sin \theta \end{aligned}\)

Now, the electric force on each charge is,

\(\begin{aligned}{c}F = k\frac{{\frac{Q}{2} \times \frac{Q}{2}}}{{{r^2}}}\\ = k\frac{{{Q^{^2}}}}{{4{r^2}}}\\ = k\frac{{{Q^{^2}}}}{{4{{\left( {2l\sin \theta } \right)}^2}}}\\ = k\frac{{{Q^{^2}}}}{{16{l^2}{{\sin }^2}\theta }}\end{aligned}\)

Consider the figure below.

Apply the equilibrium condition in the horizontal direction.

\(\begin{aligned}{c}T\sin \theta = F\\T\sin \theta = k\frac{{{Q^{^2}}}}{{16{l^2}{{\sin }^2}\theta }}\end{aligned}\) … (i)

Apply the equilibrium condition in the vertical direction.

\(T\cos \theta = mg\) … (ii)

Now, dividing equation (i) by (ii) you get,

\(\begin{aligned}{c}\frac{{T\sin \theta }}{{T\cos \theta }} = \frac{{k\frac{{{Q^{^2}}}}{{16{l^2}{{\sin }^2}\theta }}}}{{mg}}\\mg\tan \theta = k\frac{{{Q^{^2}}}}{{16{l^2}{{\sin }^2}\theta }}\\{Q^2} = \frac{{16mg{l^2}{{\sin }^2}\theta \tan \theta }}{k}\\Q = \sqrt {\frac{{16mg{l^2}{{\sin }^2}\theta \tan \theta }}{k}} \end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}Q = \sqrt {\frac{{16 \times \left( {0.021\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times {{\left( {0.78\;{\rm{m}}} \right)}^2}{{\sin }^2}{{26}^ \circ }\tan {{26}^ \circ }}}{{9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{C}}^{{\rm{ - 2}}}}}}} \\ = 4.6 \times {10^{ - 6}}\;{\rm{C}}\end{aligned}\)

Thus, the total charge that must have been applied to the electroscope is \(4.6 \times {10^{ - 6}}\;{\rm{C}}\).