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(A) The capacitance $C$ of a spherical capacitor with inner radius $r$ and outer radius $R$ is given by the formula:$C = 4 \pi \epsilon_0 \frac{rR}{R

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The capacitance $C$ of a spherical capacitor with inner radius $r$ and outer radius $R$ is given by the formula:$C = 4 \pi \epsilon_0 \frac{rR}{R - r}$Given $C = 100 \ pF = 100 	imes 10^{-12} \ F$,and the spacing $d = R - r = 1 \ cm = 0.01 \ m$.We know that $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2$,so $4 \pi \epsilon_0 = \frac{1}{9 	imes 10^9}$.Substituting the values into the formula:$100 	imes 10^{-12} = \frac{1}{9 	imes 10^9} \cdot \frac{r(r + 0.01)}{0.01}$$10^{-10} = \frac{r^2 + 0.01r}{9 	imes 10^7 	imes 0.01}$$10^{-10} = \frac{r^2 + 0.01r}{9 	imes 10^5}$$r^2 + 0.01r = 9 	imes 10^{-5}$$r^2 + 0.01r - 0.00009 = 0$Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{-0.01 \pm \sqrt{(0.01)^2 - 4(1)(-0.00009)}}{2}$$r = \frac{-0.01 \pm \sqrt{0.0001 + 0.00036}}{2}$$r = \frac{-0.01 \pm \sqrt{0.00046}}{2} \approx \frac{-0.01 \pm 0.0214}{2}$Taking the positive root: $r \approx \frac{0.0114}{2} = 0.0057 \ m = 0.57 \ cm$.Wait,re-evaluating the calculation: $C = 4 \pi \epsilon_0 \frac{r(r+d)}{d}$. For $C = 100 \ pF$,$r(r+0.01) = 100 	imes 10^{-12} 	imes 9 	imes 10^9 	imes 0.01 = 0.009$.$r^2 + 0.01r - 0.009 = 0$. Solving this: $r = \frac{-0.01 + \sqrt{0.0001 + 0.036}}{2} = \frac{-0.01 + 0.19}{2} = 0.09 \ m = 9 \ cm$.

The capacitance of a spherical capacitor is $100 \ pF$. If the spacing between the two spheres is $1 \ cm$,then the radius of the inner sphere of the capacitor is: (in $cm$)

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