Consider an infinite slab of nonconducting material that lies between x = -d/2 a
ID: 2028425 • Letter: C
Question
Consider an infinite slab of nonconducting material that liesbetween x = -d/2 and x = +d/2 (total thickness d) parallel to the yz-plane with volume
charge density ?. Use unit vector notation (ˆi, ˆj,kˆ ,rˆ) for the E fields.
(a). Find the electric field E for all values of x. (Use Gauss's law)
(b). Find the potential V for all values of x. Take the zero of potential to be at (0,d/2,0).
(c). Consider a sphere of the same nonconducting substance with volume charge
density ? and radius d/2 centered at (0,0) without the slab present. Find the electric field
for all values of r.
(d). Find the potential V everywhere. Take the zero of potential to be at (0,d/2,0).
(e). Suppose a spherical section of the material in part (a) centered at (0,0) and radius
d/2 is removed. Find the new E field everywhere. (Hint: Take a look at problem 22-83
which you had for homework)
(f). Find the potential V everywhere. Take the zero of potential to be at (0,d/2,0)
Explanation / Answer
a) Using gauss law, when x<-d/2 or x>d/2, then we treat the slab as a single parallel plane. Gauss law states that from an infinate plane, Ei = /2. Ej and Ek are zero.
When -d/2<x<d/2, we need to find the amount of charge within the slab as a function or x. So
= Q/V, and the volume of the slab is heightXlengthXwidth. So = Q / hXlXwidth
However, the Q we need is a function of of x, the width, so density X volume as a function = q as a function
( Q / hXlXwidth ) ( hXlXr ) = q
The width of the total volume is d/2. This is because when we choose the gaussian shape, it will be centered at 0,0 so as to make the math easier. After we simplify, we get:
q = Q(2r/d)
Now to use gauss law:
EA = q/.
E = Qr / Ad however, Q/A is so:
Ei = 2r / d. r is on the x axis, so you may replace r with x.
Ej and Ek are zero
b) The electric field everywhere outside the slab is constant, so if we integrate E with respect to x we get the voltage function. However, because it is constant, we can simply multiply by x.
for x<-d/2 and x>d/2, V = x/2
For inside the slab, let us intergrate the E function with respect to x ( or r ) so:
V = 2r / d dr = r2 / d
c) The electric field outside the sphere can treat the sphere as a point particle centered at the origin. Therefore, for x>d/2,
Er = kQ/r2
When inside, we use a density calculation again, this time of a sphere, and then apply gauss law. It will in turn become:
Er = 8kQr / d3
d) The voltage outside the sphere is the same as the voltage from a point particle, so it becomes:
V = kQ/r
The voltage INSIDE becomes V = E dr
V = 8kQr / d3 dr = 4kQr2/d3
cannot do e) and f) because you did not supply the additional information. For future reference, this question was WAY to big to be answered for 175 points. Most people ignore a problem of this length and complexity. I recommend breaking it up into smaller questions next time.