Answer:
(a) [tex]C_{capacitance}=1.32*10^{-11}F=13.2pF[/tex]
(b) [tex]r_{a}=0.034m=3.4cm[/tex]
(c) [tex]E=2.6*10^{4}N/C[/tex]
Explanation:
Given data
Charge q=3.3 nC
Voltage V=250.0 V
Radius r=4.70 cm=0.047 m
Required
(a) Capacitance C
(b) The Radius of inner sphere
(c) Electric field
Solution
For Part (a) Capacitance
The Capacitance is given by:
[tex]C_{capacitance}=\frac{Q_{charge}}{V_{volatge}}\\ C_{capacitance}=\frac{3.3*10^{-9}C}{250V}\\ C_{capacitance}=1.32*10^{-11}F=13.2pF[/tex]
For Part (b) The radius of inner sphere
The Capacitance of spherical co-ordinates is given by:
[tex]C=\frac{4\pi E_{o}}{\frac{1}{r_{a}} -\frac{1}{r_{b}} }\\ Simplify.it\\\frac{1}{r_{a}} -\frac{1}{r_{b}}=\frac{4\pi (8.85*10^{-12})}{1.32*10^{-11}}=8.42m^{-1}\\\frac{1}{r_{a}}=8.42+\frac{1}{r_{b}}\\\frac{1}{r_{a}}=8.42+(1/0.047) \\\frac{1}{r_{a}}=29.7\\r_{a}=1/29.7\\r_{a}=0.034m=3.4cm[/tex]
For Part (c) The electric field
The electric field according to Gauss Law is given by:
[tex]EA=\frac{Q}{E_{o}} \\E=\frac{Q}{4\pi E_{o}r_{b}^2}=\frac{kQ}{r_{a}^2} \\E=\frac{9*10^9*3.30*10^{-9}}{(0.034)^2}\\ E=2.6*10^{4}N/C[/tex]
