91影视

We were given a spherical capacitor where the radius of the first sphere is

$C=4蟺{系}_{鈭�}\frac{r_a{r}_b}{r_b-r_a}$

a) The capacitance C is the magnitude Q of the charge on either sphere divided by the potential difference $V_{ab}$between the spheres where we can get the capacitance as the terms of the radius of each sphere

$$\begin{gathered} C=4蟺{系}_{鈭�}\frac{r_a{r}_b}{r_b-r_a} \\ {r}_{b}C=r_{a}C=4蟺{系}_{鈭�}{r}_a{r}_b \\ {r}_{a}C=r_b(C-4蟺{系}_{鈭�}{r}_a) \\ {r}_a=\frac{r_{b}C}{\left(C-4蟺{系}_{鈭�}{r}_a\right)} \end{gathered}$$

Now Substitute our values for C, $r_a$into the above equation to get the radius of the outer sphere where

$$\begin{gathered} r_b=\frac{r_{a}C}{\left(C-4蟺{系}_{鈭�}{r}_a\right)} \\ =\frac{(0.15m)(116脳{10}^{-12}F)}{116脳{10}^{-12}F-4蟺\left[8.854脳{10}^{-12}F/m\right]\left[0.15m\right]} \\ =0.175m \\ =17.5cm \end{gathered}$$

The radius of the outer sphere is 17.5cm

b) The two spheres have the same magnitude of the charge Q but in the opposite sign. So, we can use the value of the capacitance and the potential difference $V_{ab}$to get the charge by applying the equation

$Q=C{V}_{ab}$

where the charge is directly proportional to the capacitance C. Now substitute our values to get the charge Q

$$\begin{gathered} Q=C{V}_{ab} \\ =(116脳{10}^{-12}F)(220.0V) \\ =25.52脳{10}^{-9}C \\ =25.52nC \end{gathered}$$

Hence the magnitude of the charge Q on each sphere is 25.52nC