## anonymous 3 years ago 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?

1. anonymous

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)

2. anonymous

Use the fact that the electric field is the same everywhere (think of the electric flux through a cylinder on the plane).

3. anonymous

Simpler when the field is constant which it is here - always for infinite plane $\Delta V = E*\Delta x$

4. anonymous

so 5 = E*Delta x find E and then it will help find$\Delta x$

5. anonymous

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.

6. anonymous

|dw:1347905879599:dw|

7. anonymous
8. anonymous

@RaphaelFilgueiras pls can you be more legible with ur formulas.....@benpen thanks for the link

9. anonymous

but the value i get is 8.98 x 10 ^ -4 m @RaphaelFilgueiras any clues

10. anonymous

it's close my is 8.85x10^-4