need help with 4, 5, and 6 An infinite sheet of charge is located in the y-z pla

Question

need help with 4, 5, and 6

An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity 1 = 0.35 C/m2. Another infinite sheet of charge with uniform charge density 2 = -0.24 C/m2 is located at x = c = 25 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 10.5 cm and x = 14.5 cm).

1)

What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (5.25 cm, 0)?

Your answer has been judged correct; the exact answer is: 33363.6858 N/C

2)

What is a, the charge density on the surface of the conducting slab at x = 10.5 cm?

Your answer has been judged correct; the exact answer is: -0.295  C/m2

3)

What is V(R) - V(P), the potentital difference between point P and point R, located at (x,y) = (5.25 cm, -14.5 cm)?

Your answer has been judged correct; the exact answer is: 0 V

4)

What is V(S) - V(P), the potential difference between point P and point S, located at (x,y) = (19.75 cm, -14.5 cm)?

Your submissions:

----------------- V

5)

What is Ex(T), the x-component of the electric field at point T, located at (x,y) = (30.25 cm, -14.5 cm)?

-------------- N/C

6)

Which of the folowing plots gives the correct x-dependence for the potential function between x = 0 and x = 25 cm?

A

B

C

D

E

F

Explanation / Answer

ginve uncharged conducting slab is placed between two infininte charges sheets

charge densities

sigma1 = 0.35 uC/m^2, at x = 0 m

sigma2 = -0.24 uC /m^2, at x = 0.25 m

slab thickness, t = 4 cm

hence 2*a + t = 25

a = 10.5 cm

a. at point P

electric field E = sigma1/2*epsilon - sigma2/2*epsilon

E1 = 33289.572394 V/m to the right

b. at x = 10.5 cm

charge density on the slab surface be sigma'

then

from gauss law

electric field inside the conductor is 0

hence

E1*A = -sigma'*A/epsilon

where A is the are of the gaussean surface

hence

sigma = -0.295 micro C/m^2

c. as electric field inside the two plates is constant hence the potential difference falls with x only and not y

now as the two points P and R are on the same x coordinate hence potential difference between the two is 0

d. between point P and S

-dV/dx = E

dV = -E*dx = -33289.572394*(5.25*2 + 4)/100

dV = 4826.98799713 V

e. at x = 30.25 cm

electric field E is given by

E = (sigma1 + sigma2)/2*epsilon = (0.35 - 0.24)*10^-6*2*pi*8.98*10^9 = 6206.53044 N/C

f. the potential function decreases linearly from x = 0 to x = 10.5 and x = 14.5 to x = 25 as electric field is pointing to the right and is constant

also, potential inside the slab is non zero but constant

hence

B is the corect answer

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