Two Infinite Non-Conducting Planes

  • Thread starter Thread starter blueskadoo42
  • Start date Start date
  • Tags Tags
    Infinite Planes
AI Thread Summary
The electric field values for the regions around two infinite non-conducting planes are given as 19j V/m above plane 'r', -31j V/m between the planes, and -19j V/m below plane 's'. To find the surface charge density (σr) for plane 'r', the correct relationship to use is E = σ/(2ε0), taking into account the direction of the electric field. The total electric field in each region is the sum of the contributions from both planes, requiring careful attention to the signs of the fields. Three equations can be set up based on the electric fields in the three regions to solve for the surface charge densities. Understanding the contributions from each plane is crucial for correctly calculating σr.
blueskadoo42
Messages
26
Reaction score
0

Homework Statement


The electric field equals 19j for points above plane 'r', -31j between the planes, and -19j below plane 's', where j is the unit vector in the +y direction and the fields are in V/m. Calculate σr, the suface charge density for plane 'r'.

positive y- straight up
positive x-to the right.


Homework Equations



E=Qenc/Epsilon0

The Attempt at a Solution



i tried adding the 19j + -31j then multiplying that by Epsilon 0. that didnt work. not sure how to add or apply the electric fields. any pointers on how to solve that?
 
Physics news on Phys.org
bump!
 
anyone? i don't know which field to add or subtract. any help?
 
You're using the wrong relationship between charge and E. While your relationship is strictly correct, what you need is

E=\frac{\sigma}{2\epsilon_0}

where E>0 means that it points away from the plane where \sigma is (and \sigma>0), and E<0 means that it points toward the plane where \sigma is (and \sigma<0). Then, you need to realize that the E in each of the three regions is the sum of the E's from the individual planes, and you have to be careful about the + and - signs. So, you can write 3 equations for the three regions in terms of the three E's and two \sigma's.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top