PhoenixFire
  • PhoenixFire
Two point Charges each with a charge of +10nC sit at (0,0,0) and (1,2,1). Find the Electric Field E at point (4,4,4). Give me a second to post my work.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
PhoenixFire
  • PhoenixFire
Using Coulombs Law: \[\vec{E}=\frac{kq}{\left| \vec{r} \right|^3}\vec{r}\] Where \(k=\frac{1}{4\pi \epsilon}=9*10^9\) The vector between the point charges and observation point: \[\vec{r_1}=P_1-P_{obs}=<-4,-4,-4>\]\[\vec{r_2}=P_2-P_{obs}=<-3,-2,-3>\] \[\left| \vec{r_1} \right| = \sqrt{48}\]\[\left| \vec{r_2} \right| = \sqrt{22}\] And then basically plugging in the values: \[\vec{E_1}=\frac{9*10*10^{9-9}}{\sqrt{48}^3}\vec{r_1}=<-1.083,-1.083,-1.083>\] \[\vec{E_2}=\frac{9*10*10^{9-9}}{\sqrt{22}^3}\vec{r_2}=<-2.617,-1.744,-2.617>\] And sum them up:\[\vec{E_{tot}}=\vec{E_1}+\vec{E_2}=<-3.7, -2.827, -3.7>\]
PhoenixFire
  • PhoenixFire
Not sure if I am doing it right or not. I have no answers so I can't verify it at the moment.
anonymous
  • anonymous
seems right

Looking for something else?

Not the answer you are looking for? Search for more explanations.