Question 26.2: An isolated conducting sphere has a 10 cm radius. One wire c......
An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a current of 1.000 000 0 A out of it. How long would it take for the sphere to increase in potential by 1000 V?
Suppose the charge on the sphere increases by Δq in time Δt. Then, in that time its potential increases by
\Delta V=\frac{\Delta q}{4 \pi \varepsilon_0 r} ,
where r is the radius of the sphere. This means \Delta q=4 \pi \varepsilon_0 r \Delta V . Now, \Delta q=\left(i_{\text {in }}-i_{\text {out }}\right) \Delta t , where i_{in} is the current entering the sphere and i_{out} is the current leaving. Thus,
\begin{aligned} \Delta t & =\frac{\Delta q}{i_{\text {in }}-i_{\text {out }}}=\frac{4 \pi \varepsilon_0 r \Delta V}{i_{\text {in }}-i_{\text {out }}}=\frac{(0.10 \,m )(1000 \,V )}{\left(8.99 \times 10^9 \,F / m \right)(1.0000020\, A -1.0000000\, A )} \\ & =5.6 \times 10^{-3}\, s . \end{aligned}