Answer to Question #151761 in Physics for Arslan khan

Question

Four identical point charges (q = + 10.0 microCoulombs) are located on the corners of a rectangle as shown in Figure P23.9. The dimensions of the rectangle are L = 60.0 cm and W = 15.0 cm. Calculate the magnitude and direction of the net electric force exerted on the charge at the lower left corner by the other three charges.

Answer 1

$F_{net}=\sqrt{F_x^2+F_y^2}$

$F_x=k\frac{qq}{L^2}+k\frac{qq}{(\sqrt{L^2+W^2})^2}\cos\alpha$

$\cos\alpha=\frac{L}{\sqrt{L^2+W^2}}=\frac{0.6}{\sqrt{0.6^2+0.15^2}}=0.9701\to \alpha=14°$

$F_x=k\frac{qq}{L^2}+k\frac{qq}{(\sqrt{L^2+W^2})^2}\cos\alpha=9\cdot10^9\frac{10\cdot10^{-6}\cdot10\cdot10^{-6}}{0.6^2}+$

$+9\cdot10^9\frac{10\cdot10^{-6}\cdot10\cdot10^{-6}}{(\sqrt{0.6^2+0.15^2})^2}\cos14°=4.78(N)$

$F_y=k\frac{qq}{W^2}+k\frac{qq}{(\sqrt{L^2+W^2})^2}\sin\alpha=9\cdot10^9\frac{10\cdot10^{-6}\cdot10\cdot10^{-6}}{0.15^2}+$

$+9\cdot10^9\frac{10\cdot10^{-6}\cdot10\cdot10^{-6}}{(\sqrt{0.6^2+0.15^2})^2}\sin14°=40.57(N)$

$F_{net}=\sqrt{4.78^2+40.57^2}=40.85(N)$ . Answer

$\tan\gamma =\frac{F_y}{F_x}=\frac{40.57}{4.78}=8.487\to \gamma\approx83°$ (relative to the negative direction of the x-axis). Answer