6. Consider a slab of material of thickness t filled with a constant negative and positi charge densities as shown in fig. 6A [a toy model for a pn material or a diode]. (a) Use Gauss Law to determine the electric field everywhere[assume the slab is infinite large in y z plane]. (b) Determine the potential difference V_t/2 – V+t/2 (c) Show that If by means of a battery , a potential difference V is held over the ent: sandwich the electric field and the potential at position x is given by[see Figure 6B]:
6. Consider a slab of material of thickness t filled with a constant negative and positi charge densities as shown in fig. 6A [a toy model for a pn material or a diode]. (a) Use Gauss Law to determine the electric field everywhere[assume the slab is infinite large in y z plane]. (b) Determine the potential difference V_t/2 – V+t/2 (c) Show that If by means of a battery , a potential difference V is held over the ent: sandwich the electric field and the potential at position x is given by[see Figure 6B]:
Related questions
Question
![(A)
(B)
n
t/2 t/2
Gaussian
surface
-P| +p
FP +p
+t/2
V.
b
-t/2
-t/2
+t/2
(C)
Current is high
[forward bias]
Current is low
[reverse bias]
p n
pn
Figure 6:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09b3997c-2096-4573-b74b-055c69181c0d%2F575f8822-3ad7-40d1-99e9-7f2286179d61%2F3udqri4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(A)
(B)
n
t/2 t/2
Gaussian
surface
-P| +p
FP +p
+t/2
V.
b
-t/2
-t/2
+t/2
(C)
Current is high
[forward bias]
Current is low
[reverse bias]
p n
pn
Figure 6:
![6. Consider a slab of material of thickness t filled with a constant negative and positive
charge densities as shown in fig. 6A [a toy model for a pn material or a diode].
(a)Use Gauss Law to determine the electric field everywhere[assume the slab is infinitely
large in y z plane].
(b) Determine the potential difference V_t/2 – V+t/2
(c) Show that If by means of a battery , a potential difference V, is held over the entire
sandwich the electric field and the potential at position x is given by[see Figure 6B]:
plt/2-|c|)
E = Eo -
€0
pt
V = V, + Eo (- – x) + elt+2=)²
p(t+2x)2
8e0
; -t/2 < x < 0
where Eo :
t
4€0
V = V, + Eo (-– x) + +
pæ(t-x)
2e0
; 0<x< t/2
8e0
6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09b3997c-2096-4573-b74b-055c69181c0d%2F575f8822-3ad7-40d1-99e9-7f2286179d61%2Fhk19e38_processed.png&w=3840&q=75)
Transcribed Image Text:6. Consider a slab of material of thickness t filled with a constant negative and positive
charge densities as shown in fig. 6A [a toy model for a pn material or a diode].
(a)Use Gauss Law to determine the electric field everywhere[assume the slab is infinitely
large in y z plane].
(b) Determine the potential difference V_t/2 – V+t/2
(c) Show that If by means of a battery , a potential difference V, is held over the entire
sandwich the electric field and the potential at position x is given by[see Figure 6B]:
plt/2-|c|)
E = Eo -
€0
pt
V = V, + Eo (- – x) + elt+2=)²
p(t+2x)2
8e0
; -t/2 < x < 0
where Eo :
t
4€0
V = V, + Eo (-– x) + +
pæ(t-x)
2e0
; 0<x< t/2
8e0
6
Expert Solution
Step 1
According to Gauss law,
If oppositely charges parallel conducting plates are treated like infinite planes then Gauss' law can be used to calculate the electric field .
(a)For conducting surface approach,
For charge sheet approach,
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