anonymous
  • anonymous
A point charge q1= -4.00nC is at the point x= 0.60m , y= 0.80m , and a second point chargeq2 = +6.00nC is at the point x= 0.60m , y= 0. Calculate the magnitude of the net electric field at the origin due to these two point charges.
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
note that \(\vec E_{total} = \vec E_{q1} + \vec E_{q2}\) where \(\vec E = \frac{Q}{4 \pi \epsilon_o r^2}(\cos \theta \vec a_x + \sin \theta \vec a_y) \) where for q1, \(r_1^2 =(|\vec r_1|)^2 =(0.6^2 +0.8^2)\), \(\theta_1 =\tan^{-1} \frac{0.8}{0.6}\) where for q2, \(r_1^2 =(|\vec r_2|)^2 =(0.6^2 +0^2)\), \(\theta_1 =\tan^{-1} \frac{0}{0.6}\) do you need further help?
anonymous
  • anonymous
Thank you for help. I wasnt able to reply for some reason
anonymous
  • anonymous
you're welcome :)

Looking for something else?

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