Answer to Question #143193 in Physics for Maria

Question

A −1.00-nC point charge is at the origin, and a +4.00-nC
point charge is on the

Answer 1

a)

$$E(3)=(9\cdot10^{9})\frac{(4\cdot10^{-9})}{(3-2)^2}-(9\cdot10^{9})\frac{(1\cdot10^{-9})}{(3-0)^2}=35\frac{N}{C}$$

b)

$$E(1)=-(9\cdot10^{9})\frac{(4\cdot10^{-9})}{(1-2)^2}-(9\cdot10^{9})\frac{(1\cdot10^{-9})}{(1-0)^2}\\=-45\frac{N}{C}$$

c)

$$E(-2)=-(9\cdot10^{9})\frac{(4\cdot10^{-9})}{(-2-2)^2}+(9\cdot10^{9})\frac{(1\cdot10^{-9})}{(-2-0)^2}\\=0\frac{N}{C}$$