Asked by Annoyingmous

We have an infinite, non-conducting sheet of negligible thickness carrying a uniform surface charge density +σ=7.00e-6 C/m2 and, next to it, an infinite parallel slab of thickness D=17 cm with uniform volume charge density ρ=−4.94e-5 C/m3 (see sketch). Note that the slab has a negative charge. All charges are fixed in place and cannot move.

Reminder: The "x.xxenn C/m2 and C/m3" notations for σ and ρ, respectively, mean "x.xx×10nn C/m2 and C/m3".


Assuming the infinite sheet is in the x-y plane, and zˆ points as shown in the figure, calculate the components of the electric field at the locations listed below. Since we do not ask for the direction separately, these components could be positive or negative.

Give all of your answers in Volts/m.

(a) a distance h=4 cm above (i.e. in the +zˆ direction) the positively charged sheet.
Ex



Ey



Ez


(b) inside the slab at a distance d=4 cm below (i.e. in the −zˆ direction) the positively charged sheet. Note that d<D so that this is also a distance 13 cm above the bottom of the slab.
Ex



Ey



Ez

(c) a distance H=29 cm below (i.e. in the −zˆ direction) the charged sheet. Note that H>D so that this is also a distance 12 cm below the bottom of the slab.


Ex



Ey



Ez

Answered by ^_ ^
a) 0
0
sigma/(2*epsilon_0) - rho*D/(2*epsilon_0)

b)
0
0

c)
0
0
Answered by robbo
what about c part 3 and b part 3???
Answered by demon
c part 3)
-sigma/(2*epsilon_0)-rho*(D-
d)/(2*epsilon_0) +rho*d/(2*epsilon_0)
Answered by demon
sry it was b part 3)
Answered by robbo
demon, any help with the capacitor brushes problem?
Answered by me
anyone for question 7
Answered by AAA
a) (sigma/(2*epsilon_0))-(rho*D/(2*epsilon_0))

c)-sigma/(2*epsilon_0)+(rho*D/(2*epsilon_0))
Answered by AAA
Please, Question 3
Answered by AAA
b) -sigma/(2*epsilon_0)-rho*(D-
d)/(2*epsilon_0) +rho*d/(2*epsilon_0)
Answered by Anonymous
Please question 7, 9, 10 and 11.
Answered by j
anyone for 8
Answered by OVNI
Please 7, 9, 10 and 11.
Answered by AAA
7) a) lambda/0.5
b) lambda/0.125

Please question 3, 5, 8
Answered by boss
For Inductor
a: I=0 A
b: (Supply Voltage/R)*(1-e^(-Rt/L))
c: see above (put the value of t in sec

For Capacitor
Use v(t)=Supply Voltage*(1-e^(-t/RC))
Divide the above answer by R to get the Current

Answered by boss
question 9
a: Ec1=Voltage/plate separation
b: because the battery is connected the E field does not change

c: when the battery is disconnected then
Q1+Q2 must equal Q1'+Q2'
Ec1=Q1'/(k*C1*d) and Ec2=Q2'/(C2*d)
where Q1'=(Q1+Q2)/(1+(C2/K*C1)
and Q2'=C2/(K*C1)*(Q1+Q2)/(1+(C2/K*C1))
There are no AI answers yet. The ability to request AI answers is coming soon!