Answer:
Therefore,
[tex]V_{10}=2.34\times 10^{6}\ V[/tex]
[tex]V_{20}=1.17\times 10^{6}\ V[/tex]
[tex]V_{14}=1.67\times 10^{6}\ V[/tex]
Explanation:
Given:
A spherical conductor has a radius of 14.0 cm
Q = 26.0 μC ( Assume it to be microC in the question it is miliC )
To Find:
Electric potential at
(a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm from the center.
Solution:
Electric potential due to point charge Q at any distance r from the charge is,
[tex]V=\dfrac{kQ}{r}[/tex]
Where,
V = Electric Potential
k = Coulombs constant = 9 × 10⁹ Nm²/C².
Q = Charge
r = distance in meter
Substituting the values we get
At r = 10 cm = 0.1 m
[tex]V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.1}[/tex]
[tex]V=2.34\times 10^{6}\ V[/tex]
At r = 20 cm = 0.2 m
[tex]V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.2}[/tex]
[tex]V=1.17\times 10^{6}\ V[/tex]
At r = 14 cm = 0.14 m
[tex]V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.14}[/tex]
[tex]V=1.67\times 10^{6}\ V[/tex]
Therefore,
[tex]V_{10}=2.34\times 10^{6}\ V[/tex]
[tex]V_{20}=1.17\times 10^{6}\ V[/tex]
[tex]V_{14}=1.67\times 10^{6}\ V[/tex]
