An infinite charged sheet has a surface charge density of 1 x 10^-7 Cm^-2 .How far apart are the equipotential surfaces whose potentials differ by 5volts?

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

An infinite charged sheet has a surface charge density of 1 x 10^-7 Cm^-2 .How far apart are the equipotential surfaces whose potentials differ by 5volts?

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Work out the electric field at any point, and use the fact that potential energy is \[-\int\limits_{\infty}^{a}F(x)dx\](i.e. a unit charge is dragged from infinity, parallel to the field lines, to a)
Use the fact that the electric field is the same everywhere (think of the electric flux through a cylinder on the plane).
Simpler when the field is constant which it is here - always for infinite plane \[\Delta V = E*\Delta x\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

so 5 = E*Delta x find E and then it will help find\[\Delta x\]
Yes, basically my integral simplifies to \[U=-Fx\] As F is not actually a function of x. Evaluated from infinity to a, or more simply the change in PE (from b to a), yields Mikael's result.
|dw:1347905879599:dw|
http://www.youtube.com/watch?v=ldJhMDuOGxY
@RaphaelFilgueiras pls can you be more legible with ur formulas.....@benpen thanks for the link
but the value i get is 8.98 x 10 ^ -4 m @RaphaelFilgueiras any clues
it's close my is 8.85x10^-4

Not the answer you are looking for?

Search for more explanations.

Ask your own question