Question

What is the magnitude of the electric field at a distance of 1.5m from a point charge with q=3.5C?

Answer 1

We find the magnitude of the electric field with $\vert{\vec{E_i}}\vert=\dfrac{kq}{d^2}$, where the constant k = 9 X 10⁹ Nm²/C². Since the distance d = 1.5 m, and the charge q = 3.5 C, we proceed to substitute to find the magnitude of E:

$\vert\vec{E}\vert=\dfrac{(9×10^9\frac{Nm^2}{C^2})(3.5\,C)}{(1.5\,m)^2}=1.4×10^{10}\frac{N}{C}$

In conclusion, the magnitude of the electric field will be |E| = 1.4 X 10¹⁰ N/C.