To find the resistance of a material, we use the formula:

[ R = frac{L}{sigma A} ]

where ( R ) is the resistance in ohms (( Omega )), ( L ) is the length of the material in meters (m), ( sigma ) is the conductivity in siemens per meter (S/m), and ( A ) is the cross-sectional area in square meters (( m^2 )). Given that conductivity is provided in millimhos per square meter, we first need to convert it to siemens per meter (S/m). Note that 1 mho is equivalent to 1 siemens (S), and therefore 1 millimho = ( 1 times 10^{-3} ) S.

Given values:

– Conductivity (( sigma )) = 2 millimhos/m(^2) = ( 2 times 10^{-3} ) S/m(^2)

– Length (( L )) = 10 m

– Cross-sectional area (( A )) = 50 m(^2)

Substituting the values into the formula:

[ R = frac{10}{(2 times 10^{-3}) times 50} ]

[ R = frac{10}{0.1} ]

[ R = 100 , Omega ]

Therefore, the resistance of the material

To calculate the capacitance of a parallel plate capacitor in air, you can use the formula:

[C = epsilon_r epsilon_0 frac{A}{d}]

Where:

– (C) is the capacitance in farads (F),

– (epsilon_r) is the relative permittivity of the dielectric material between the plates (for air, it can be approximated as 1),

– (epsilon_0) is the vacuum permittivity, equal to approximately (8.854 times 10^{-12} text{F/m}) (farads per meter),

– (A) is the area of one of the plates in square meters (m²),

– (d) is the distance between the plates in meters (m).

Given:

– (A = 20) units (assuming units means square meters, (20 m^2)),

– (d = 5 m),

The equation with given values:

[C = 1 times 8.854 times 10^{-12} text{F/m} times frac{20}{5}]

[C = 8.854 times 10^{-12} text{F/m} times 4]

[C = 35.416 times 10^{-12} text{F}]

[C = 35.416 text{pF}]

To find the charge density at a given point due to a potential function in a spherical coordinate system, we use the relation provided by the Poisson equation for electrostatics in vacuum or air, which is

[ nabla^2 V = -frac{rho}{epsilon_0} ]

Where ( nabla^2 ) is the Laplace operator (in this case, in spherical coordinates), (V) is the potential, ( rho ) is the charge density, and ( epsilon_0 ) is the permittivity of free space ((8.854 times 10^{-12} , C^2/N cdot m^2 )).

Given, ( V = frac{10sin(theta)cos(phi)}{r} )

First, let’s apply the Laplacian in spherical coordinates to (V), remembering that (r), (theta), and (phi) are the radius, polar angle, and azimuthal angle, respectively. The Laplacian of a scalar field (V(r, theta, phi)) in spherical coordinates is given by:

[ nabla^2 V = frac{1}{r^2} frac{partial}{partial r} left( r^2 frac{partial V}{partial r} right) + frac{1}{r^2 sin

The potential of a function ( V = frac{60cos(theta)}{r} ) at a point ( P(r, theta, z) ) is calculated directly by plugging the coordinates of the point into the function. Given the point ( P(3, 60, 25) ), the values of ( r ) and ( theta ) are provided in the coordinates of the point, where ( r = 3 ) and ( theta = 60 ) degrees.

First, we need to convert the angle ( theta ) from degrees to radians because trigonometric functions in calculus are typically evaluated in radians. ( theta = 60^circ = frac{pi}{3} ) radians.

Then, we can substitute ( r ) and ( theta ) into the formula:

[ V = frac{60cos(theta)}{r} = frac{60cos(frac{pi}{3})}{3} ]

Knowing that ( cos(frac{pi}{3}) = frac{1}{2} ), we get:

[ V = frac{60 times frac{1}{2}}{3} = frac{30}{3} = 10 ]

So, the potential ( V ) of the function at the point ( P(3, 60, 25) ) is ( 10