Physics, 20.09.2019 16:20 haylieshaven1

Density, density, density. (a) a charge -312e is uniformly distributed along a circular arc of radius 5.70 cm, which subtends an angle of 71o. what is the linear charge density along the arc? (b) a charge -312e is uniformly distributed over one face of a circular disk of radius 3.70 cm. what is the surface charge density over that face? (c) a charge -312e is uniformly distributed over the surface of a sphere of radius 4.20 cm. what is the surface charge density over that surface? (d) a charge -312e is uniformly spread through the volume of a sphere of radius 4.00 cm. what is the volume charge density in that sphere?

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Answer from: Jgrant7485

(a): Linear charge density of the circular arc = $\rm -7.07\times 10^{-16}\ C/m.$

(b): Surface charge density of the circular arc = $\rm -1.16\times 10^{-14}\ C/m^2.$

(c): Volume charge density of the sphere = $\rm -1.86\times 10^{-13}\ C/m^3.$

Explanation:

Part (a):

Given:

Total charge on the circular arc, $\rm q = -312\ e.$ Radius of the circular arc, $\rm r = 5.70\ cm = 0.057\ m.$ Angle subtended by the circular arc, $\rm \theta=71^\circ=71\times \dfrac{\pi }{180}\ rad = 1.23918\ rad.$

We know, e is the elementary charge whose value is $\rm 1.6\times 10^{-19}\ C.$

Therefore, $\rm q=-312\times1.6\times 10^{-19}=-4.992\times 10^{-17}\ C.$

Also, the length l of a circular arc is given as:

$\rm l = r\theta =0.057\times 1.23918=7.06\times 10^{-2}\ m.$

The linear charge density $\lambda$ of the arc is defined as the charge per unit length of the arc.

$\rm \lambda = \dfrac{q}{l} = \dfrac{-4.992\times 10^{-17}}{7.06\times 10^{-2}}=-7.07\times 10^{-16}\ C/m.$

Part (b):

Given:

Total charge on the circular disc, $\rm q = -312\ e.$ Radius of the circular disc, $\rm r = 3.70\ cm = 0.037\ m.$

$\rm q=-312\times1.6\times 10^{-19}=-4.992\times 10^{-17}\ C.$

Surface area of the circular disc, $\rm A = \pi r^2 = \pi \times (0.037)^2 = 4.3\times 10^{-3}\ m^2.$

The surface charge density $\sigma$ of the disc is defined as the charge per unit area of the disc.

$\rm \sigma = \dfrac qA=\dfrac{-4.992\times 10^{-17}}{4.3\times 10^{-3}}=-1.16\times 10^{-14}\ C/m^2.$

Part (c):

Given:

Total charge on the sphere, $\rm q = -312\ e.$ Radius of the sphere, $\rm r = 4.00\ cm = 0.04\ m.$

$\rm q=-312\times1.6\times 10^{-19}=-4.992\times 10^{-17}\ C.$

Volume of the sphere, $\rm V = \dfrac 43 \pi r^3 = \dfrac 43 \times \pi \times 0.04^3 = 2.68\times 10^{-4}\ m^3.$

The volume charge density $\rho$ of the sphere is defined as the charge per unit volume of the sphere.

$\rm \rho = \dfrac qV = \dfrac{-4.992\times 10^{-17}}{ 2.68\times 10^{-4}}= -1.86\times 10^{-13}\ C/m^3.$

Answer from: Quest

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hope this

Answer from: Quest

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