Question

What is the electric field at a location <-0.3, -0.1, 0> m, due to a particle with charge 2 nC located at the origin?

Answer 1

Step-by-step explanation:

Given

Position vector of location
`\vec{r}=-0.3x\hat{i}-0.1y\hat{j}`

Magnitude of position vector
`|\vec{r}|=√((-0.3)^2+(-0.1)^2+0)`

`|\vec{r}|=0.316\ m`

The electric field at position vector
`\vec{r}` is given by

`\vec{E}=(kq)/(r^3)\cdot \vec{r}`

`\vec{E}=(9* 10^9* 2* 10^(-9))/(0.316^2)\cdot (-0.3x\hat{i}-0.1y\hat{j})`

`\vec{E}=(180)\cdot (-0.3x\hat{i}-0.1y\hat{j})`

`\vec{E}=-54x\hat{i}-18y\hat{j}`

Electric field is

<-54, -18, 0 >