Question

determine the magnitude and direction of electric field due to the point of 4.0 to the left of a point charge of -2.5nc

Answer 1

The magnitude of the electric field is given as follows:

$$E = k\dfrac{q}{r^2}$$

where $k = 9\times 10^9N\cdot m^2/C, q = 2.5\times 10^{-6}C, r = 4m$. Thus, obtain:

$$E = 9\times 10^9\cdot \dfrac{2.5\times 10^{-9}}{4^2} \approx 5.6V/m$$

The direction is toward the charge i.e. to the right.

Answer. 5.6 V/m to the right.