Question

Two charges of -6.25nC and -12.5nC are placed 25cm in a line.
Determine the electric field at a point 10cm to the left of -6.25nC and
the electric field at a point 10cm to the left of the -12.5nC

Answer 1

$$E_1=\frac{kq}{r^2}=-\frac{9\times10^9\times12.5\times10^{-9}}{({35\times10^{-2}})^2}=-918.36N/c$$

$$E_2=\frac{kq}{r^2}=-\frac{9\times10^9\times6.5\times10^{-9}}{({10\times10^{-2}})^2}=-5850N/c$$ $$E=E_1+E_2=-918.36-5858=-6768.36N/c$$

$$E'_1=\frac{kq}{r^2}=-\frac{9\times10^9\times12.5\times10^{-9}}{({10\times10^{-2}})^2}=-11250N/c$$

$$E'_2=\frac{kq}{r^2}=-\frac{9\times10^9\times6.25\times10^{-9}}{({15\times10^{-2}})^2}=-2500N/c$$

$$E'=E'_1+E'_2=-11250-2500=-13750N/c$$