Respuesta :
Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
[tex]V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V[/tex]
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ = [tex]\frac{d}{dp}.a_p + \frac{d}{p*dQ}*a_Q + \frac{d}{dz}*a_z[/tex]
Hence,
[tex]E = \frac{100*cos(Q)}{z^2 + 1}.a_p+ \frac{100*sin(Q)}{z^2 + 1}.a_Q + \frac{-200*p*zcos(Q)}{(z^2 + 1)^2}.a_z[/tex]
Plug in the values for point P:
[tex]E = \frac{100*cos(60)}{5}.a_p+ \frac{100*sin(60)}{5}.a_Q + \frac{-200*3*2*cos(60)}{(5)^2}.a_z\\\\E = - 10*a_p +17.32 *a_Q-24*a_z[/tex]
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
[tex]a_N = - vector(E) / (dV/dN)\\\\a_N = 10 /31.2 *a_p - 17.32/31.2 *a_Q + 24/31.2 *a_z\\\\a_N = 0.32 *a_p - 0.55 *a_Q + 0.77 *a_z[/tex]
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3
The values at P(Potential Field) would be:
a). [tex]V = 30V[/tex]
b). [tex]E = -10[/tex] × [tex]a_{p}[/tex] [tex]+ 17.32[/tex] ×[tex]a_{Q} - 24[/tex] × [tex]a_{z}[/tex]
c). [tex]E = 31.241 N/C[/tex]
d). [tex]dV / dN = | E | = 31.2 N/C[/tex]
e). [tex]a_N[/tex] = [tex]0.32[/tex] × [tex]a_{p} - 0.55[/tex] × [tex]a_{Q} + 0.77[/tex] × [tex]a_{z}[/tex]
f). [tex]p_v[/tex] [tex]= 276 pC / m^3[/tex]
Given that,
[tex]V = 100/(z^2+1) rho cos[/tex] φ[tex]V[/tex]
φ [tex]= 60[/tex]°, point P at [tex]rho = 3m[/tex],
[tex]z = 2m[/tex]
a). To find the potential function of Volt at the P point, We will find out the position P's value,
[tex]V = 100/(z^2+1) rho cos[/tex] φ[tex]V[/tex]
[tex]V = (100[/tex] × [tex]3[/tex] × [tex]Cos(60))[/tex]/[tex](2^2 + 1)[/tex]
[tex]= 150/5= 30V[/tex]
b). To find Electric Field(E),
[tex]E =[/tex] - ∀.[tex]V[/tex]
Here, ∀ signifies a del function i.e.
∀ [tex]= \frac{d}{dp}.a_{p} + \frac{d}{p * dQ} * aQ + d/dz * a_{z}[/tex]
By applying the values associated with P, we get
[tex]E = -10[/tex] × [tex]a_{p}[/tex] [tex]+ 17.32[/tex] ×[tex]a_{Q} - 24[/tex] × [tex]a_{z}[/tex]
c). Electric Field's Magnitude,
[tex]E = \sqrt{ ((-10)^2 + (17.32)^2 + (24)^2)}[/tex]
[tex]E = \sqrt{975.9824\\}[/tex]
[tex]E = 31.241 N/C[/tex]
d). The field strength of E equals to:
| - ∀.V |
∵
e). [tex]a_N[/tex] denoting unit vector in direction of [tex]dV/dN[/tex]:
[tex]a_{N} = -vector (E)/(dV/dN)[/tex]
[tex]a_N[/tex] = [tex]0.32[/tex] × [tex]a_{p} - 0.55[/tex] × [tex]a_{Q} + 0.77[/tex] × [tex]a_{z}[/tex]
f). The density of charge in free space:
[tex]p_v[/tex] = E.∈[tex]_o[/tex]
∵ [tex]= 8.85*10^-12 p_v = 31.24.8.85*10^-12 p_v = 276 pC / m^3[/tex]
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