## woleraymond 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. henpen

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. henpen

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

3. Mikael

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

4. Mikael

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

5. henpen

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. RaphaelFilgueiras

|dw:1347905879599:dw|

7. henpen
8. woleraymond

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

9. woleraymond

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

10. RaphaelFilgueiras

it's close my is 8.85x10^-4