Respuesta :
Answer:
(a). The capacitance is [tex]6.43\times10^{-12}\ F[/tex]
(b). The potential difference is necessary to produce these charges on the cylinders are 2.41 V.
Explanation:
Given that,
Charge = 15.5 pC
Inner Radius = 0.600 mm
Outer radius = 5.20 mm
Length = 25.0 cm
(a). We need to calculate the capacitance
Using formula of capacitor
[tex]C=\dfrac{2\pi\epsilon_{0}l}{ln\dfrac{b}{a}}[/tex]
Put the value into the formula
[tex]C=\dfrac{2\pi\times8.85\times10^{-12}\times25.0\times10^{-2}}{ln\dfrac{5.20}{0.600}}[/tex]
[tex]C=6.43\times10^{-12}\ F[/tex]
The capacitance is [tex]6.43\times10^{-12}\ F[/tex]
(b). We need to calculate the potential difference is necessary to produce these charges on the cylinders
Using formula of charge
[tex]q=CV[/tex]
[tex]V=\dfrac{q}{C}[/tex]
Put the value into the formula
[tex]V=\dfrac{15.5\times10^{-12}}{6.43\times10^{-12}}[/tex]
[tex]V=2.41\ V[/tex]
Hence, (a). The capacitance is [tex]6.43\times10^{-12}\ F[/tex]
(b). The potential difference is necessary to produce these charges on the cylinders are 2.41 V.
