Question

A circular disk has surface charge density 24 nc/cm2 what will the surface charge density be if the radius of the disk is doubled?

Answer 1

The surface charge density of a circular disk is given by the following relationship:

σ =
`(Q)/(\pi*R^(2))` -- (A)

Where,

σ = Surface charge density.
Q = Charge on the surface
R = Radius of the circular disk.

Now according to the above equation, we can infer:
1) Surface charge density σ is directly proportional to the Charge Q on the surface of the disk.
2) Surface charge density σ is inversely proportional to the square of Radius R of the circular disk.

As the charge is 24nC on the circular disk, and there is no evidence of charge being changed in the question, I would assume that the charge is constant.

Let's apply the condition:
If R = 2R(if the radius of the disk is made doubled)

Plug in the value in (A):
A => σ(new) =
`(Q)/(\pi*(2R)^(2))`

Therefore,
σ(new) =
`(Q)/(\pi*4*R^(2))`

Hence,
σ(new) = (1/4)σ

Conclusion:
If the Radius of the disk is doubled, the new surface charge density would become one-fourth of the old surface charge density. Hence,

σ(new) =
`( (1)/(4))( (24nC)/(cm^(2)) )`

Ans: σ(new) = 6nC/
`cm^(2)`

-i

Answer 2

Answer: 6 nC / cm^2

Justification:

1) Keeping the same charge (constant charge) the charge density is inversely realted to the surface.

=> charge density = A / surface

where A is a constant

2) Surface is prortional to square of the radius, so the charge density is inversely related to the square of the radius:

=> charge density = B / (radius)^2

where B is a constant

3) this shows that if the radius is doubled the density falls to 1/4:

charge density = B / (2* radius)^2 = B / [ 4*(radius)^2 ] = (1/4) B / (radius)^2

Conclusion: when the radius is doubled the density charge falls to one quarter.

=> [24 nC/cm^2 ] / 4 => 6nC / cm^2.