Question

An infinite line charge produces a field of $9\times10^{4}\,\text{N/C}$ at a distance of $2\,\text{cm}$. Calculate the linear charge density.

Hint:

For an infinite line charge E = lambda / (2 pi epsilon_0 r). Rearrange to lambda = E times 2 pi epsilon_0 r with r = 0.02 m.

Answer 1

Step 1: The electric field at a perpendicular distance \(r\) from an infinite line charge of linear density \(\lambda\) is:
\[E = \frac{\lambda}{2\pi\epsilon_0\,r}\]
Step 2: Rearrange to find the linear charge density:
\[\lambda = E\times 2\pi\epsilon_0\,r\]
Step 3: Substitute \(E = 9\times10^{4}\,\text{N C}^{-1}\), \(r = 2\,\text{cm} = 0.02\,\text{m}\), and \(\epsilon_0 = 8.85\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\text{m}^{-2}\):
\[\lambda = (9\times10^{4})\times 2\pi\,(8.85\times10^{-12})\times(0.02)\]
Step 4: Do the arithmetic:
\[\lambda = 1.0\times10^{-7}\,\text{C m}^{-1}\]
\[\boxed{\lambda = 1.0\times10^{-7}\,\text{C m}^{-1} = 0.1\,\mu\text{C m}^{-1}}\]

Answer 2

Gauss's law with a cylinder (derivation route).
Step 1: Wrap the line charge in a coaxial cylinder of radius \(r\) and length \(L\). The charge enclosed is \(\lambda L\). Only the curved side contributes to flux:
\[E\,(2\pi r L) = \frac{\lambda L}{\epsilon_0}\]
Step 2: The \(L\) cancels, leaving the field formula. Solve for \(\lambda\):
\[\lambda = 2\pi\epsilon_0\,r\,E\]
Step 3: Using \(\dfrac{1}{2\pi\epsilon_0} = 2k = 1.8\times10^{10}\), write \(\lambda = \dfrac{E\,r}{2k}\) and substitute:
\[\lambda = \frac{(9\times10^{4})\times(0.02)}{1.8\times10^{10}}\]
Step 4: Evaluate:
\[\lambda = \frac{1.8\times10^{3}}{1.8\times10^{10}} = 1.0\times10^{-7}\,\text{C m}^{-1}\]
\[\boxed{\lambda = 1.0\times10^{-7}\,\text{C m}^{-1}}\]