Question about conducting plates

[erik_8991]

Homework Statement

Two infinite sheets with surface charge density σ1 and σ2, respectively, are oriented perpendicular to the x-axis. An infinite, conducting slab of thickness a is placed between the charged sheets as shown in the figure. The conducting plate has a net charge per unit area of σC.

Homework Equations

Calculate the x-component of the net electric field at x = 0.

(a) Ex = +7.85 × 105 V/m
(b) Ex = +6.22 × 105 V/m
(c) Ex = +5.65 × 105 V/m
(d) Ex = -2.67 × 105 V/m
(e) Ex = -3.85 × 105 V/m


QUESTION 15**
Calculate the surface charge density σR induced on the right hand face of the conducting plate.

(a) σR = +3.75 μC/m2
(b) σR = +2.00 μC/m2
(c) σR = -1.50. μC/m2
(d) σR = -3.00 μC/m2
(e) σR = -4.25 μC/m2

The Attempt at a Solution

14. for this number, i suppose left part of the conducting plate has the charge density -1.5. that gives the total E at x = 0 equal to 10. i use the formula E = sigma/2Eo = 10/(2*8.85) to get the answer C

15. I have no idea. I thought it was c, but the answer is B.

thank you so much!

 

[Andrew Mason]

Apply Gauss' law to the Gaussian cylinder of cross-sectional area da enclosing the left surface of the conducting sheet (ie. ending in the middle of the conducting sheet where the field = 0) and x=0. You know that the net field through the left surface of area da is equal to the field at x=0 (ie. the answer to question 14) and the field through the right surface of area da is 0.

So applying Gauss' law: E_x\cdot da = dq/\epsilon_0 = \sigma_L da/\epsilon_0

which means that \sigma_L = E_x\epsilon_0 (work out the polarity from the direction of the field, which is the direction in which a positive charge would move).

Then it is just a matter of working out what the charge density on the right is so that the total charge density is \sigma_c = -3.0 \mu C

AM

 

[ideasrule]

14. for this number, i suppose left part of the conducting plate has the charge density -1.5. that gives the total E at x = 0 equal to 10. i use the formula E = sigma/2Eo = 10/(2*8.85) to get the answer C

You don't need to suppose. The electric field induced by an infinite sheet is sigma/e0, regardless of the distance to the sheet. That means the conducting plate induces sigma_c/e0 everywhere except inside the plate, so you can just add up the electric fields and get the anser.