Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One, with charge per unit area o, is positioned at r=-d, and the other, with charge per unit area 20, is positioned at r +d. Fig. I +20 (a) Find the electrie field vector E for all values of z. Now an electrically neutral, infinite conducting slab of thickness d is inserted in between the charged sheets, as shown below in Fig. 2. +20 Fig. 2
Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One, with charge per unit area o, is positioned at r=-d, and the other, with charge per unit area 20, is positioned at r +d. Fig. I +20 (a) Find the electrie field vector E for all values of z. Now an electrically neutral, infinite conducting slab of thickness d is inserted in between the charged sheets, as shown below in Fig. 2. +20 Fig. 2
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![Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One,
with charge per unit area o, is positioned atr= -d, and the other, with charge per unit area
20, is positioned at r = +d.
y
Fig. I
+20
(a) Find the electric field vector E for all values of z.
Now an electrically neutral, infinite conducting slab of thickness d is inserted in between
the charged sheets, as shown below in Fig. 2.
+20
Fig. 2
*- +d2 -+d
(b) Find the location and values of any induced charge distributions in or on the slab.
(c) Recompute the electric field vector E for all values of z for this new configuration.
(d) Plot the clectric potential V(r) on the x-axis, for all values of r for this configuration.
Set the zero of the electric potential at r = 0, i.e. take V(r = 0) = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6bd502c-7281-4dca-bc00-207e3be2d405%2F54930064-877a-4892-a35c-6f5cfd5f441b%2Fyuui8ir_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider two infinite insulating sheets of charge as shown edge-on in Fig. 1 below. One,
with charge per unit area o, is positioned atr= -d, and the other, with charge per unit area
20, is positioned at r = +d.
y
Fig. I
+20
(a) Find the electric field vector E for all values of z.
Now an electrically neutral, infinite conducting slab of thickness d is inserted in between
the charged sheets, as shown below in Fig. 2.
+20
Fig. 2
*- +d2 -+d
(b) Find the location and values of any induced charge distributions in or on the slab.
(c) Recompute the electric field vector E for all values of z for this new configuration.
(d) Plot the clectric potential V(r) on the x-axis, for all values of r for this configuration.
Set the zero of the electric potential at r = 0, i.e. take V(r = 0) = 0.
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