Question

The sides of a right triangle are BC=3 cm, AC=4cm, and AB=5cm. If +10 uC is placed at the corners B and C, a) what is the magnitude and direction of the electric field at A?

Answer 1

$E=k\frac{q}{x^2}$

$E_1=k\frac{q}{x_1^2}=9\cdot10^9\cdot \frac{10\cdot10^{-6}}{0.04^2}=56.25\cdot10^6\ (V/m)$

$E_2=k\frac{q}{x_2^2}=9\cdot10^9\cdot \frac{10\cdot10^{-6}}{0.05^2}=36\cdot10^6\ (V/m)$

$E=\sqrt{E_1^2+E_2^2+2E_1E_2\cos\alpha}=$

$=\sqrt{(56.25\cdot10^6)^2+(36\cdot10^6)^2+2\cdot 56.25\cdot10^6\cdot 36\cdot10^6\cdot(4/5)}=$

$=87.75\cdot10^6\ (V/m)$ . Answer