Question

A +4.5 PC charge is located at coordinates x= +3.0 cm and y = +3.0 cm. Another +4.5 uc charge is located at coordinates x= -3.0 cm and y = +3.0 cm. What is the magnitude of the electric field at the origin? A. 6.75 x 104 V/m B. 2.35 x 104 V/m C. 1.73 x 104 V/m D. 6.75 x 103 V/m E. 13.5 x 10 V/m

Answer 1

Here, the components of the electric field along the Y-axis(the sine component) will add and the the component along the X-axis will cancel out. We can see that the angle the distance vector make is

$\tan^{-1}\frac{0.03}{0.03}=45^\circ$

The distance vector's magnitude is

$\\ r=\sqrt{0.03^2+0.03^2} \\ r=0.0424\mathrm{m}$

Putting all this into our formula, we get

$\\ E=k\frac{2Q\sin45}{r^2} \\ E=k\frac{2\times (4.5\times 10^{-6})\sin 45}{1.8\times 10^{-3}} \\ E=31.81\times 10^6\mathrm{V/m}$

None of the options match the correct answer.