Question

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?

Answer 1

(a): Linear charge density of the circular arc =
`\rm -7.07* 10^(-16)\ C/m.`

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

(c): Volume charge density of the sphere =
`\rm -1.86* 10^(-13)\ C/m^3.`

Step-by-step 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* (\pi )/(180)\ rad = 1.23918\ rad.`

We know, e is the elementary charge whose value is
`\rm 1.6* 10^(-19)\ C.`

Therefore,
`\rm q=-312*1.6* 10^(-19)=-4.992* 10^(-17)\ C.`

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

`\rm l = r\theta =0.057* 1.23918=7.06* 10^(-2)\ m.`

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

`\rm \lambda = (q)/(l) = (-4.992* 10^(-17))/(7.06* 10^(-2))=-7.07* 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*1.6* 10^(-19)=-4.992* 10^(-17)\ C.`

Surface area of the circular disc,
`\rm A = \pi r^2 = \pi * (0.037)^2 = 4.3* 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 = \frac qA=(-4.992* 10^(-17))/(4.3* 10^(-3))=-1.16* 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*1.6* 10^(-19)=-4.992* 10^(-17)\ C.`

Volume of the sphere,
`\rm V = \frac 43 \pi r^3 = \frac 43 * \pi * 0.04^3 = 2.68* 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 = \frac qV = (-4.992* 10^(-17))/( 2.68* 10^(-4))= -1.86* 10^(-13)\ C/m^3.`