Help with #4 (in V) and #5 (N/C) .. Thanks! An infinite sheet of charge is locat
ID: 1656394 • Letter: H
Question
Help with #4 (in V) and #5 (N/C) .. Thanks! An infinite sheet of charge is located in the y-z plane at x 0 and has uniform charge density | = 0.58 pC/m2. Another infinite sheet of charge with uniform charge density 2--0.38 uc/m2 is located at x = c-29 cm. An uncharged infinite conducting slab is placed halfway in between these sheets i.e between x = 12.5 cm and x = 16.5 cm) a/2 1) What is E(P), the x-component of the electric field at point P, located at (x,y) (6.25 cm, 0)? 54237.29 N/C Submit 2) What is the charge density on the surface of the conducting slab at x = 12.5 cm? 48 pC/m2 | Submit 3) What is V(R) V(P), the electrical potentital difference between point P and point R, located at (x,y) (6.25 cm, 16.5 cm)? 0 Submit 4) What is V(S) V(P), the potentital difference between point P and point S, located at (x,y) (22.75 cm, -16.5 cm)? Submit 5) What is E(T), the x-component of the electric field at point T, located at (x,y) (35.25 cm, -16.5 cm)?Explanation / Answer
given infinite sheet at x = 0, charge density sigma1 = 0.58*10^-6 C/m^2
given infinite sheet at x = 29, charge density sigma2 = -0.38*10^-6 C/m^2
given infinite conductor slab , uncharged from x = 12.5 cm to x = 16.5 cm
1. at x = 6.25 cm
let electric field = E
now, we know that electrc field at a distance x from an infinitely charged plate is E = sigma/2*epsilon ( where epsilon is permittivity of free space)
hence electric field at x = 6.25 cm is (sigma1 - sigma2)/2*epsilon = 0.96*10^-6*8.98*10^9*2pi = 54166.08389 V/m
2. At x = 12.5 cm, let the slab have charge density sigma
then for field to be 0 inside the conductor
(0.58 + 0.38 + sigma) = 0
sigma = -0.48 micro C/m^2
3. as electric field is same for P and R, and P and R both lie on equipotential surface
V(P) - V(R) = 0 V
4. V(s) - V(P) = V(S) - V(R)
now, electric field is from left to right, so V(S) < V(P)
and, V(S) - V(P) = - 54166.08389 * (0.2275 - 0.0625) = -8937.403 V
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