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Language:
English
Chapter 7
2.
3. 1
CHAPTER 7 DRILLS
D7.1
(a) V |P (1,2,3) =
4(2)(3)
(1)2
+1 = 12 V
As, ρv = −∇2
V, so we first calculate ∇2
V :
∇V = −8
yzx
(x2
+1)2
+
4z
x2
+1 +
4y
x2
+1
⇒ ∇2
V = 32 yzx2
(x2
+1)3 −8 yz
(x2
+1)2
⇒ ∇2
V |P (1,2,3) = 12;
so, ρv = −∇2
V = −o(12) = −106.25pC
m3
(b) V
|
P (3, π
3
,2) =
−
22.5 V
As, ρv = −∇2
V, so we first calculate ∇2
V :
∇2
V = 20 cos (2φ) −20 cos(2 φ)
⇒ ∇2
V |P (3, π
3
,2) = 0;
so, ρv = −∇2
V = −o(0) = 0 pC
m3
Solved by Zaeem Ahm ad Vara ich
Com piled by www.EngineersTeam.tk
4. (c) V |P (0.5,45o
,60o
) = 4 V
As, ρv = −∇2
V, so we first calculate ∇2
V :
∇2
V = 4 cos(φ)
r4 −2 cos(φ)
r4
(sin(θ))2
⇒ ∇2
V |P (0.5,45o
,60o
) = 0;
so, ρv = −∇2
V = −o(0) = 0 pC
m3
D7.2
Apply the formulae & concepts to find the answers!
www.engineersteam.tk
Compiled by www.EngineersTeam.tk
2
D7.3
(a) The solution to Laplace’s equation 1
ρ
∂
∂ρ
ρ ∂
∂ρ
= 0 is:
V = A ln ρ + B
Putting the given values of V & ρ and solving the simultaneous equations, we get:
A = −73.9 & B = 101.28
so, V = −73.9 ln ρ + 101.28
1 1
Compiled by www.EngineersTeam.tk
5. Link
3 of 4
Now, E = −∇V = ρ
(73.9) aρ = √
10
(73.9) aρ = 23.36aφ
(
∵
ρ =
√
3
2
+ 1
2
)
so, |E| = 23.36 V
m
(b) The solution to Laplace’s equation 1
ρ2
∂2
V
∂φ2 = 0 is:
V = Aφ + B
Putting the given values of V & φ and solving the simultaneous equations, we get:
A = −85.9 & B = 64.9
so, V = −85.9φ + 64.9
Now, E = −∇V = 1
ρ
(85.9) aφ = 1√
10
(85.9) aφ = 27.16 aφ
(∵ ρ = √32 + 12)
so, |E| = 27.16 V
m
D7.4 & D7.5
Not included in the course!
www.engineersteam.tk
Compiled by www.EngineersTeam.tk
6. 3
D7.6
The solution to this problem depends on how you proceed in each iteration and on your initial estimate.
The dotted lines show how the initial estimate was found.
(a) 20.8 V
(b) 44.47 V
(c) 89.8 V
Compiled by www.EngineersTeam.tk
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