Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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?
Thank you for help. I wasnt able to reply for some reason
you're welcome :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question