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Q63PE

Expert-verifiedFound in: Page 666

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) In Figure 18.59, four equal charges \(q\) lie on the corners of a square. A fifth charge \(Q\) is on a mass \(m\) directly above the center of the square, at a height equal to the length \(d\) of one side of the square. Determine the magnitude of \(q\) in terms of \(Q\), \(m\), and \(d\), if the Coulomb force is to equal the weight of \(m\). (b) Is this equilibrium stable or unstable? Discuss.**

**Figure 18.59 Four equal charges on the corners of a horizontal square support the weight of a fifth charge located directly above the center of the square.**

(a) The magnitude of the charge \(q\) is \(0.459\frac{{mg{d^2}}}{{KQ}}\).

(b) The system is in stable equilibrium.

**According to Coulomb, when two charges are separated by some distance, the tries attract or repel the other charge with some force known as Coulomb force or electrostatic force****.** The expression for the Coulomb force is,

\(F = \frac{{KqQ}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant, \(q\) and \(Q\) are the charges separated by a distance \(r\).

The weight of charge \(Q\) is,

\({F_g} = mg\)

Here, \(m\) is the mass of the charge \(Q\), and \(g\) is the acceleration due to gravity.

Due to symmetry of the system, the horizontal components of the forces will cancel each other leaving only the vertical components of the force.

The force on the charge \(Q\) is represented as,

Force on the charge \(Q\)

The distance of the charge \(Q\) from the charge \(q\) is,

\(\begin{array}{c}r = \sqrt {{{\left( {\frac{d}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{d}{{\sqrt 2 }}} \right)}^2} + {d^2}} \\ = \sqrt {\frac{3}{2}} d\end{array}\)

The Coulomb force between the charges \(Q\) and \(q\) is,

\(\begin{array}{c}F = \frac{{KQq}}{{{r^2}}}\\ = \frac{{KQq}}{{{{\left( {\sqrt {\frac{3}{2}} d} \right)}^2}}}\\ = \frac{{2KQq}}{{3{d^2}}}\end{array}\)

The vertical component of the force is,

\(\begin{array}{c}{F_y} = F\sin \theta \\ = \frac{{2KQq}}{{3{d^2}}}\sin \theta \end{array}\)

From geometry,

\(\begin{array}{c}\sin \theta = \frac{d}{r}\\ = \frac{d}{{\sqrt {\frac{3}{2}} d}}\\ = \sqrt {\frac{2}{3}} \end{array}\)

Therefore,

\({F_y} = \frac{{2KQq}}{{3{d^2}}} \times \sqrt {\frac{2}{3}} \)

Due to symmetry the net force on the charge \(Q\) is,

\({F_n} = 4{F_y}\)

Substituting all known values,

\(\begin{array}{c}{F_n} = 4 \times \frac{{2KQq}}{{3{d^2}}} \times \sqrt {\frac{2}{3}} \\ = 2.177\frac{{KQq}}{{{d^2}}}\end{array}\)

Since, the electrostatic force supports the weight of the charge \(Q\). Therefore,

\(\begin{array}{c}{F_n} = {F_g}\\2.177\frac{{KQq}}{{{d^2}}} = mg\end{array}\)

Rearranging the above expression in order to get an expression for the charge \(q\),

\(\begin{array}{c}q = \frac{{mg{d^2}}}{{2.177KQq}}\\ = 0.459\frac{{mg{d^2}}}{{KQ}}\end{array}\)

Hence, the magnitude of the charge \(q\) is \(0.459\frac{{mg{d^2}}}{{KQ}}\).

Horizontal movement of the charge will result in a net horizontal component to restore the mass back to its original position.

Hence, the system is in stable equilibrium.

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