(a) Calculate the net x -component of the electric fieldat the following positio

Question

(a) Calculate the net x-component of the electric fieldat the following positions:

At x = -2 cm: Ex = N/C   

At x = 2 cm: Ex = N/C    

HELP:  Do not use Gauss's Law, usesuperposition.

At x = 4.5 cm: Ex = N/C    *
0   OK

HELP:  Apply the general principlethat applies for the electric field anywhere inside a conductor atequilibrium.

At x = 10 cm: Ex = N/C    
-4.52e-5   NO

HELP:  Do not use Gauss's Law, usesuperposition.
HELP:  Make repeated use of thealgebraic form of the electric field due to a single infinite sheetof uniform surface charge.

(b) Calculate the surface charge densities on the left-hand(sa) and right-hand(sb) faces of theconducting slab.You may also find it useful to note therelationship between sa and sb.

sa = C/m2    

sb = C/m2    

An infinite sheet of charge, oriented perpendicular to thex-axis, passes through x = 0. It has area density sigma^2 = -2MuC/m^2. A thick, infinite conducting slab, alsooriented perpendicular to the x-axis, occupies the regionbetween x = a and x = b, wherea = 4 cm and b = 5 cm. The conducting slab has anet charge per unit area of sigma2 = 5 MuC/m^2. (a) Calculate the net x-component of the electric fieldat the following positions: At x = -2 cm: Ex = N/C At x = 2 cm: Ex = N/C HELP: Do not use Gauss's Law, usesuperposition. At x = 4.5 cm: Ex = N/C * 0 OK HELP: Apply the general principlethat applies for the electric field anywhere inside a conductor atequilibrium. At x = 10 cm: Ex = N/C -4.52e-5 NO HELP: Do not use Gauss's Law, usesuperposition. HELP: Make repeated use of thealgebraic form of the electric field due to a single infinite sheetof uniform surface charge. (b) Calculate the surface charge densities on the left-hand(sa) and right-hand(sb) faces of theconducting slab.You may also find it useful to note therelationship between

Explanation / Answer

a) 1 = -2 µC/m2, 2= 5 µC/m2, a = 4 cm, b = 5cm. for the slab, total charge Q = 22A, where A is thearea for one surface Q = (a + b)*A, soa + b =22The electric field due to one infinite plateis /(2o) At x = -2 cm: Ex =-1/(2o) - (a +b)/(2o) = -1/(2o) -22/(2o) = -4.52*105N/C At x = 2 cm: Ex =1/(2o) - (a +b)/(2o) = 1/(2o) -22/(2o) = -6.78*105N/C At x = 4.5 cm: Ex = 0 (inside theconductor) At x = 10 cm: Ex =1/(2o) + (a +b)/(2o) = 1/(2o) +22/(2o) = 4.52*105N/C b) At x = 4.5 cm: Ex =1/(2o) + (a -b)/(2o) = 0 (inside theconductor) 1 + a - b =0 a - b = -1 = 2C/m2 on the other hand, a + b = 22 = 10C/m2solve the above 2 equations andgeta = 6 C/m2 b = 4 C/m2

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