Respuesta :
Answer:
(A). The capacitance is [tex]15.2\times10^{-12}\ F[/tex]
(B). The radius of the inner sphere is 0.0327 m .
(C). The electric field just outside the surface of the inner sphere is [tex]2.9458\times10^{4}\ V/m[/tex]
Explanation:
Given that,
Charge = 3.50 nC
Potential difference = 230.0 V
Radius = 4.30 cm
(a). We need to calculate the capacitance
Using formula of capacitance
[tex]C=\dfrac{Q}{V}[/tex]
[tex]C=\dfrac{3.50\times10^{-9}}{230.0}[/tex]
[tex]C=15.2\times10^{-12}\ F[/tex]
(b). We need to calculate the radius of the inner sphere
Using formula of capacitance
[tex]C=\dfrac{4\pi\epsilon_{0}r_{a}r_{b}}{r_{b}-r_{a}}[/tex]
Put the value into the formula
[tex]15.2\times10^{-12}=4\pi\times8.85\times10^{-12}(\dfrac{r_{a}\times4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}})[/tex]
[tex]\dfrac{r_{a}4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}}=\dfrac{15.2\times10^{-12}}{4\pi\times8.85\times10^{-12}}[/tex]
[tex]\dfrac{r_{a}4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}}=0.13667[/tex]
[tex]r_{a}\times4.30\times10^{-2}=0.13667\times(4.30\times10^{-2}-r_{a})[/tex]
[tex]r_{a}(4.30\times10^{-2}+0.13667)=0.13667\times4.30\times10^{-2}[/tex]
[tex]r_{a}=\dfrac{0.13667\times4.30\times10^{-2}}{(4.30\times10^{-2}+0.13667)}[/tex]
[tex]r_{a}=0.0327\ m[/tex]
(c). We need to calculate the electric field just outside the surface of the inner sphere
Using formula of electric filed
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{Q}{r_{a}^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times3.50\times10^{-9}}{0.0327}[/tex]
[tex]E=2.9458\times10^{4}\ V/m[/tex]
Hence, (A). The capacitance is [tex]15.2\times10^{-12}\ F[/tex]
(B). The radius of the inner sphere is 0.0327 m .
(C). The electric field just outside the surface of the inner sphere is [tex]2.9458\times10^{4}\ V/m[/tex]
