Question

Four point charges, \(q,\) are fixed to the four corners of a square that is \(10.0 \mathrm{~cm}\) on a side. An electron is suspended above a point at which its weight is balanced by the electrostatic force due to the four electrons, at a distance of \(15.0 \mathrm{nm}\) above the center of the square. What is the magnitude of the fixed charges? Express the charge both in coulombs and as a multiple of the electron's charge.

Short answer

Based on the problem, we found out that the magnitude of the fixed charges at the corners of the square is \(2.10 \times 10^{-12}\,\mathrm{C}\), which is approximately \(1.31 \times 10^7\) times the electron's charge.

Step-by-step solution

Write down givens

The distance between the center of the square and the electron is \(15.0 \mathrm{nm}\). The side of the square is \(10.0 \mathrm{cm}\). The charge of the electron is \(-e\), where \(e = 1.6 \times 10^{-19} \mathrm{C}\), and the mass of the electron is \(m_e = 9.11 \times 10^{-31} \mathrm{kg}\). The gravitational acceleration is \(g = 9.81 \mathrm{m/s^2}\).

Calculate the distance between the electron and each charge

Let's denote the distance from the electron to each corner charge as \(r\). Since the electron is directly above the center of the square, we can use the Pythagorean theorem to find \(r\): $$ r = \sqrt{\left(\frac{10.0 \textrm{ cm}}{2}\right)^2 + \left(\frac{10.0 \textrm{ cm}}{2}\right)^2 + (15.0 \mathrm{nm})^2}. $$ Convert lengths to meters first: $$ r = \sqrt{\left(\frac{0.1}{2}\right)^2 + \left(\frac{0.1}{2}\right)^2 + (15.0 \times 10^{-9})^2}. $$ Now we find \(r\): $$ r = \sqrt{(0.05^2 + 0.05^2 + 2.25 \times 10^{-16})\,\mathrm{m^2}} = 7.08 \times 10^{-2}\,\mathrm{m}. $$

Find the weight of the electron

The weight \(W\) of the electron can be obtained by multiplying its mass, \(m_e\), by the gravitational acceleration, \(g\): $$ W = m_e \cdot g = (9.11 \times 10^{-31}\,\mathrm{kg}) (9.81\, \mathrm{m/s^2}) = 8.93 \times 10^{-30}\,\mathrm{N}. $$

Apply Coulomb's law to find the total electrostatic force

The total electrostatic force \(F\) acting on the electron is the vector sum of the forces due to the four charges, which are all equal in magnitude (\(F_{ij} = \frac{k|q_iq_j|}{r_{ij}^2}\), \(i, j = 1, 2, 3, 4\) and \(i \neq j\)). Since the charges are situated symmetrically, the horizontal components of the forces will cancel out. So we will consider only the vertical components of each force. The electrostatic force on the electron due to one charge \(q\) has a vertical component: $$ F_{ij}^{(v)} = F_{ij} \cos \theta = \frac{k|qe|}{r^2} \cos \theta, $$ where \(\theta\) can be found using the sine rule in the triangle with sides \(r\), \(0.1\), and \(0.1\sqrt{2}\) m: $$ \sin \theta = \frac{0.1\sqrt{2}}{2r}. $$ Then $$ \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{15.0 \times 10^{-9}}{r}. $$ We substitute this back to obtain the expression for the vertical component of the electrostatic force due to one fixed charge: $$ F_{ij}^{(v)} = \frac{k|qe|}{r^2} \cdot \frac{15.0 \times 10^{-9}}{r}. $$

Equate the weight of the electron and the total electrostatic force

The total vertical electrostatic force acting on the electron is four times the force due to one charge, so $$ \sum F_{ij}^{(v)} = 4 \cdot F_{ij}^{(v)} = 4 \cdot \frac{k|qe|}{r^2} \cdot \frac{15.0 \times 10^{-9}}{r}. $$ Equating this to the weight of the electron, we can solve for \(q\): $$ 8.93 \times 10^{-30}\,\mathrm{N} = 4 \cdot \frac{(8.99 \times 10^9\, \mathrm{N \cdot m^2/C^2})(1.6 \times 10^{-19}\, \mathrm{C})|q|}{(7.08 \times 10^{-2}\, \mathrm{m})^2} \cdot \frac{15.0 \times 10^{-9}}{7.08 \times 10^{-2}\, \mathrm{m}}. $$ Now simplify and solve for \(|q|\): $$ |q| = \frac{8.93 \times 10^{-30}\, \mathrm{N}(7.08 \times 10^{-2}\, \mathrm{m})^2}{4(8.99 \times 10^9\, \mathrm{N \cdot m^2/C^2})(1.6 \times 10^{-19}\, \mathrm{C})(15.0 \times 10^{-9})} = 2.10 \times 10^{-12}\,\mathrm{C}. $$

Express the charge both in coulombs and as a multiple of the electron's charge

We found the magnitude of the charge to be \(|q| = 2.10 \times 10^{-12}\, \mathrm{C}\). To express it as a multiple of the electron's charge, we divide by \(e\): $$ |q|/e = \frac{2.10 \times 10^{-12}\, \mathrm{C}}{1.6 \times 10^{-19}\, \mathrm{C}} = 1.31 \times 10^7. $$ Therefore, the magnitude of the fixed charges is \(2.10 \times 10^{-12}\,\mathrm{C}\) or around \(1.31 \times 10^7\) times the electron's charge.