Question

Density, density, density. (a) A charge -301e is uniformly distributed along a circular arc of radius 5.60 cm, which subtends an angle of 53o. What is the linear charge density along the arc? (b) A charge -301e is uniformly distributed over one face of a circular disk of radius 4.80 cm. What is the surface charge density over that face? (c) A charge -301e is uniformly distributed over the surface of a sphere of radius 5.20 cm. What is the surface charge density over that surface? (d) A charge -301e is uniformly spread through the volume of a sphere of radius 4.10 cm. What is the volume charge density in that sphere?

Answer 1

Answer:

(a): Linear charge density of the circular arc = $\rm -9.03\times 10^{-16}\ C/m.$

(b): Surface charge density of the circular arc = $\rm -6.65\times 10^{-15}\ C/m^2.$

(c): Volume charge density of the sphere =  $\rm -1.67\times 10^{-13}\ C/m^3.$

Explanation:

Part (a):

Given:

• Total charge on the circular arc, $\rm q = -301\ e.$

• Radius of the circular arc, $\rm r = 5.60\ cm = 0.056\ m.$

• Angle subtended by the circular arc, $\rm \theta=53^\circ=53\times \dfrac{\pi }{180}\ rad = 0.952\ rad.$

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

Therefore, $\rm q=-301\times1.6\times 10^{-19}=-4.816\times 10^{-17}\ C.$

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

$\rm l = r\theta =0.056\times 0.952=5.33\times 10^{-2}\ m.$

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

$\rm \lambda = \dfrac{q}{l} = \dfrac{-4.816\times 10^{-17}}{5.33\times 10^{-2}}=-9.03\times 10^{-16}\ C/m.$

Part (b):

Given:

• Total charge on the circular disc, $\rm q = -301\ e.$

• Radius of the circular disc, $\rm r =4.80\ cm = 0.048\ m.$

$\rm q=-301\times1.6\times 10^{-19}=-4.816\times 10^{-17}\ C.$

Surface area of the circular disc, $\rm A = \pi r^2 = \pi \times (0.048)^2 = 7.24\times 10^{-3}\ m^2.$

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

$\rm \sigma = \dfrac qA=\dfrac{-4.816\times 10^{-17}}{7.24\times 10^{-3}}=-6.65\times 10^{-15}\ C/m^2.$

Part (c):

Given:

• Total charge on the sphere, $\rm q = -301\ e.$

• Radius of the sphere, $\rm r = 4.10\ cm = 0.041\ m.$

$\rm q=-301\times1.6\times 10^{-19}=-4.816\times 10^{-17}\ C.$

Volume of the sphere, $\rm V = \dfrac 43 \pi r^3 = \dfrac 43 \times \pi \times 0.041^3 = 2.89\times 10^{-4}\ m^3.$

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

$\rm \rho = \dfrac qV = \dfrac{-4.816\times 10^{-17}}{ 2.89\times 10^{-4}}= -1.67\times 10^{-13}\ C/m^3.$