Question

Three equal charges are placed on three corners of a square. If the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{b}}\) has magnitude \(F_{\mathrm{ba}}\) and the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{c}}\) has magnitude \(F_{\mathrm{ca}},\) what is the ratio of \(F_{\mathrm{ca}}\) to $F_{\mathrm{ba}} ?$

Short answer

Answer: The ratio is \(\frac{1}{2}\).

Step-by-step solution

Understand the square and charges

Consider three corners of a square and denote the charges on the corners as \(Q_a\), \(Q_b\), and \(Q_c\). Let's assume the side length of the square is \(s\). Keep in mind that the charges are equal, and we want to find the ratio \(F_{ca}/F_{ba}\).

Apply Coulomb's Law to find force magnitudes

Coulomb's Law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the distance squared between them: \(F = k \frac{q_1 \cdot q_2}{r^2}\) where \(F\) is the magnitude of the force, \(k\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges. In our case, \(q_1=q_2=Q\) for all the charges.

Calculate the distance between the charges

To find the ratio \(F_{ca}/F_{ba}\), we must first find the distances between the charges. Since the charges are placed on the corners of a square, we know that: \(r_{ab} = s\) \(r_{ac} = s \sqrt{2}\) (by using the Pythagorean theorem)

Find the magnitudes of the forces

Now that we have the distances, we can find the magnitudes of the forces between the charges: \(F_{ba} = k \frac{Q \cdot Q}{(s)^2} = k \frac{Q^2}{s^2}\) \(F_{ca} = k \frac{Q \cdot Q}{(s \sqrt{2})^2} = k \frac{Q^2}{2s^2}\)

Calculate the ratio \(F_{ca}/F_{ba}\)

Divide the magnitudes to find the desired ratio: \(\frac{F_{ca}}{F_{ba}} = \frac{k \frac{Q^2}{2s^2}}{k \frac{Q^2}{s^2}} = \frac{1}{2}\) The ratio of \(F_{ca}\) to \(F_{ba}\) is \(\frac{1}{2}\).