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Three very large sheets are separated by equal distances of 15.0 cm. The first a

ID: 2157812 • Letter: T

Question

Three very large sheets are separated by equal distances of 15.0 cm. The first and third sheets are very thin and nonconducting and have charge per unit area of +5.00mu C/m3 and +5.00 mu C/m3 respectively The middle sheet is conducting but has no net charge. What is the electric field inside the middle sheet? What is the electric field between the left and middle sheets? What is the electric field between the middle and the right sheets? What is the induced electric charge density on the left and right surfaces (interface) of the middle sheet?

Explanation / Answer

(a)Since the middle sheet is a conductor so the electric field inside it is zero.

the charge density induced on the left and right hand side of the middle sheet will be same in magnitude and opposite in sign. So, their electric fields cancel outside the middle sheet as the electric fields due to these two sides will be opposite in direction and same in magnitude.

(b) Thus, the field in left and middle sheet has only net two contributions. One from the left sheet and another from the right sheet. the field of theses two is same in magnitude (as the charge density is same) and same in direction and is directed towards right.

So the field is (E_{lm}=2* rac{5*10^{-6}}{2epsilon_0}=10*10^{-6}/(2*8.85*10^{-12})=5.65*10^{5}N/m)

(c) Similar to case (b) the electric fields from the two sides of middle sheet cancel and the electric field is same in magnitude and direction as in (b)

(d) Let the induced charge density be on the left side of middle sheet, then charge density induce on right side will be -

electric field inside the middle sheet should be zero. And total electric field will be

(E_{m}= rac{5*10^{-6}}{2epsilon_0}+ rac{sigma}{2epsilon_0}- rac{-sigma}{2epsilon_0}- rac{-5*10^{-6}}{2epsilon_0}=0)

thus, (sigma=-5 mu C/m^2)

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