Question

A pair of very big, parallel sheets of metal carrying opposite charges of equal magnitude are separated by 2.20 cm.

A) If the surface charge density for each sheet has magnitude 47.0 nanoC/m^2, what is the magnitude of E in the region between them? (Answer in N/C)

B) What is the potential difference between the two sheets of metal?

Answer 1

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm.?

(a)If the surface charge density for each plate has magnitude 47.0 nC/m2, what is the magnitude of E_field in the region between the plates?
(b)What is the potential difference between the two plates?
(c)If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field?
(d)If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the potential difference?

(a) |E| = [{47*(10?

Answer 2

Electric field E   = surface charge density/eo

E   = sigma /eo

E = 47e-9/.(8.85e-12)

E = 5.31*10^3 N/C
-----------------------------------------

E = V/d

so PD V = Ed

V = 5.31*10^3* 0.022

V = 116.83 Volts