Answer
(a) $-3.0\times 10^7\frac{N}{C}\hat{x}$
(b) $5.9\times 10^7\frac{N}{C}\hat{x}$
Work Step by Step
(a) We know that
$\vec{E}=\frac{Kq_1}{r_1^2}(-\hat{x})+\frac{Kq_2}{r_2^2}(\hat{x})$
$\implies \vec{E}=K(\frac{-q_1}{r_1^2}+\frac{q_2}{r_2}^2)\hat{x}$
We plug in the known values to obtain:
$\implies \vec{E}=(8.99\times 10^9)(\frac{-6.2\times 10^{-6}}{(0.040)^2}+\frac{9.5\times 10^{-6}}{(0.140)}^2)\hat{x}=-3.0\times 10^7\frac{N}{C}\hat{x}$
(b) We know that
$\vec{E}=\frac{Kq_1}{r_1^2}(\hat{x})+\frac{Kq_2}{r_2^2}(\hat{x})$
$\implies \vec{E}=K(\frac{q_1}{r_1^2}+\frac{q_2}{r_2}^2)\hat{x}$
We plug in the known values to obtain:
$\implies \vec{E}=(8.99\times 10^9)(\frac{-6.2\times 10^{-6}}{(0.040)^2}+\frac{9.5\times 10^{-6}}{(0.060)^2})\hat{x}=5.9\times 10^7\frac{N}{C}\hat{x}$
