Question

A point charge q = -6.0nC is located at the origin. The electric field (in N/C) vector at the point x = -8.0m, y= +1.5m is? ​

Answer 1

Step-by-step explanation:

The position between point charge q to x = -8.0 m and y = 1.5 m is:

`r=\sqrt{x^(2)+y^(2)}=\sqrt{(-8)^(2)+(1.5)^(2)}=√(66.25)\approx 8.13` meter

Then the magnitude of the electric field E is:

`E=(1)/(4\pi\epsilon_(o)) (q)/(r^(2))=(9* 10^(9)) (6.0* 10^(-9))/(66.25)\approx 0.81 N/C`

For the vector of E:

`\tan\theta=(1.5)/(-8)=-0.1875 \rightarrow \theta = -10.61^(0)`

`E_(x)=E\cos (-10.16)=(0.81)(0.984)=0.79704 N/C`

`E_(y)=E\sin\theta = -0.81(0.184)=-0.14904 N/C`

`\vec{E}=0.79704\hat{i}-0.14904\hat{j}` in N/C