**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:

b)

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

E = - ∀.V

Where, ∀ is a del function which denoted:

∀ =

Hence,

Plug in the values for point P:

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:

**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