Question

A circular disk has a surface charge density $8nC/cm2$. What will the surface charge density be if the radius of the disk is doubled?

Short answer

The surface charge density if the radius of the disk is doubled from $8nC/c{m}^2\text{is}2\times {10}^{4}nC/m^2$.

Step-by-step solution

Given information 

$$\begin{gathered} ChargeQ=Q \\ Initialradius{r}_1=r \\ Charge{Q}_i=Q \\ Initialsurfacechargedensity{\eta }_i=8\mathrm{nC}/{\mathrm{cm}}^2 \end{gathered}$$

Explanation

Surface charge density$\eta =\frac{\text{charge}}{area}$

$$\begin{gathered} n_i=\frac{Q}{r^2} \\ \Rightarrow 8\times {10}^4 \\ \Rightarrow {\eta }_f=\frac{Q}{\mathrm{\pi }4{\mathrm{r}}^2} \\ \Rightarrow \frac{n_f}{8\times {10}^4}=\frac{{\displaystyle \frac{Q}{\mathrm{\pi }4{\mathrm{r}}^2}}}{{\displaystyle \frac{Q}{{\mathrm{\pi r}}^2}}} \\ \Rightarrow {\eta }_f=\frac{8\times {10}^4}{4} \\ \Rightarrow {\eta }_f=2\times {10}^{4}nC/m^2 \end{gathered}$$