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3 years ago

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Given the potential field in cylindrical coordinates, V = 100/ (z2+1) rho cos φV, and point P at rho = 3m, φ = 60°, z = 2m, find

values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space

2 answers:

Lisa [10]3 years ago

6 0

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Margaret [11]3 years ago

4 0

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:

$V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V$

b)

To compute the Electric field from Volt potential we have the following relation:

E = - ∀.V

Where, ∀ is a del function which denoted:

∀ = $\frac{d}{dp}.a_p + \frac{d}{p*dQ}*a_Q + \frac{d}{dz}*a_z$

Hence,

$E = \frac{100*cos(Q)}{z^2 + 1}.a_p+ \frac{100*sin(Q)}{z^2 + 1}.a_Q + \frac{-200*p*zcos(Q)}{(z^2 + 1)^2}.a_z$

Plug in the values for point P:

$E = \frac{100*cos(60)}{5}.a_p+ \frac{100*sin(60)}{5}.a_Q + \frac{-200*3*2*cos(60)}{(5)^2}.a_z\\\\E = - 10*a_p +17.32 *a_Q-24*a_z$

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:

$a_N = - vector(E) / (dV/dN)\\\\a_N = 10 /31.2 *a_p - 17.32/31.2 *a_Q + 24/31.2 *a_z\\\\a_N = 0.32 *a_p - 0.55 *a_Q + 0.77 *a_z$

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

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