Question

Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x = -2 m and has a surface charge density of ϝ = -3.2 µC/m2. The other is at x = 3 m and has a surface charge density of ϝ = 4.0 µC/m2. Find the electric field for the following locations.

(a) x < -2 m = N/C (i hat)

(b) -2 m < x < 3 m = N/C (i hat)

(c) x > 3 m = N/C (i ha)t

Answer 1

USING THE THEORY GIVEN BELOW(?₀ = 8.854 187 817... x 10^?12 C^2/Nm² the permitivity of vacuum)

a) field due to sheet at -2 id directed in +i hat (since -ve charge density) and field due to sheet at 3 is directed in -i hat (since +ve charge density)

vector E = (1/2_^o)(4-3.2)*10^-6 -i hat N/C

so

vector E = (0.4/_^o)*10^-6 -i hat N/C

b) now in between the plates plate at -2 having -ve charge density will generate E in -i hat and plate at +3 will also generate E in the -i hat (since +ve charge density so E away from plate)

so E= (1/2_^o)(4+3.2)*10^-6  -i hat N/C

vector E = (3.6/_^o)*10^-6   -i hat N/C

c)

now in RHS of the plates plate at -2 having -ve charge density will generate E in -i hat and plate at +3 will also generate E in the -i hat (since +ve charge density so E away from plate)

so E= (1/2_^o)(4-3.2)*10^-6 i hat N/C

vector E = (0.4/_^o)*10^-6 i hat N/C

Gauss's Law Derivation:

• From symmetry the E-field is always perpendicular to the sheet of charge. At any point the components of the E-field parallel to the plane add to zero because the charge is spread uniformly around the charge.
• The best choice for a Gaussian surface is a cylinder whose axis is perpendicular to the plane.

• Applying Gauss's Law to the cylinder,