Question

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

a. Calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is:_______
b. Calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest.

Answer 1

The surface charge density on a conductor can be calculated using the electric field and the permittivity of free space. Without the specific radius of curvature, we cannot compute the density for part (a), but for the smallest radius of curvature, we use the higher electric field of 74.0 kN/C.

Step-by-step explanation:

The electric field on the surface of a conductor is related to the surface charge density by the equation E = σ/ε0, where E is the electric field, σ is the surface charge density, and ε0 is the permittivity of free space. For an irregularly shaped conductor, the surface charge density σ varies depending on the radius of curvature of the surface. At points where the radius of curvature is smaller, the surface charge density is higher. To calculate the local surface charge density (σ) at a point on the surface, you multiply the electric field (E) at that point by the permittivity of free space (ε0).

For part (a), you would need the specific radius of curvature to calculate the local surface charge density, which is not provided in the question. For part (b), since the electric field is stronger where the radius of curvature is smallest, you would use the greater electric field value of 74.0 kN/C for the calculation. Using the provided equation, σ = Eε0, and taking the permittivity of free space as ε0 = 8.854 x 10-12 C2/N·m2, the calculation would be σ = 74.0 kN/C × 8.854 x 10-12 C2/N·m2.

Answer 2

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Step-by-step explanation:

Given that,

Electric field
`E_(1)=74.0\ kN/C`

Electric field
`E_(2)=14.0\ kN/C`

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

`\sigma=\epsilon_(0)E_(2)`

Put the value into the formula

`\sigma=8.85*10^(-12)*14*10^(3)`

`\sigma=1.239*10^(-7)\ C/m^2`

`\sigma=123.9*10^(-9)\ C/m^2`

`\sigma=123.9\ nC/m^2`

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

`\sigma=\epsilon_(0)E_(1)`

Put the value into the formula

`\sigma=8.85*10^(-12)*74*10^(3)`

`\sigma=6.549*10^(-7)\ C/m^2`

`\sigma=654.9*10^(-9)\ C/m^2`

`\sigma=654.9\ nC/m^2`

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².