Questions: Consider the arrangement of point charges in the figure. Part (a) Determine the direction of the force on q (with q>0 ) in the figure, given that qd=qb=+7.50 μC and qc=qd=-7.50 μC. Straight downward Correct! Part (b) Calculate the magnitude of the force on the charge q in newtons, given that the square is 13 cm on a side and q=1.25 μC. F=0.00111

Solution Steps

Let's consider the force on charge \(q\) due to each of the other charges.

• Charges \(q_a\) and \(q_b\) are positive and equal in magnitude. Since they are equidistant from \(q\), their forces on \(q\) are equal in magnitude and opposite in direction (along the diagonal). Thus, they cancel each other out.
• Charges \(q_c\) and \(q_d\) are negative and equal in magnitude. They are equidistant from \(q\), so their forces on \(q\) are equal in magnitude and directed towards them. These forces are also along the diagonal, directed downwards.

Since the forces from \(q_a\) and \(q_b\) cancel, the net force on \(q\) is due to \(q_c\) and \(q_d\). Both these forces are directed downwards. Therefore, the net force on \(q\) is directed straight downwards.

Let \(a\) be the side of the square (13 cm = 0.13 m). The distance \(r\) between \(q\) and \(q_c\) (or \(q_d\)) is half the diagonal of the square: \(r = \frac{\sqrt{2}a}{2}\). The magnitude of the force \(F_c\) due to \(q_c\) on \(q\) is given by Coulomb's law: \(F_c = k\frac{|q||q_c|}{r^2} = k\frac{|q||q_c|}{(\frac{\sqrt{2}a}{2})^2} = k\frac{2|q||q_c|}{a^2}\), where \(k = 8.988 × 10^9 Nm^2/C^2\).

Since the forces due to \(q_c\) and \(q_d\) are equal in magnitude and in the same direction, the net force \(F\) is twice the magnitude of \(F_c\): \(F = 2F_c = k\frac{4|q||q_c|}{a^2}\) Substituting the given values: \(F = (8.988 × 10^9) \frac{4 × (1.25 × 10^{-6}) × (7.50 × 10^{-6})}{(0.13)^2}\) \(F \approx 2.003 \, N\)

Final Answer