Answer:
Part a)
tex]V = -1 \times 10^8 Volts[/tex]
Part b)
[tex]V_{inner} = V_{outer} = -1 \times 10^8 volts[/tex]
Part c)
[tex]V = -7.30 \times 10^8 Volts[/tex]
Explanation:
Part a)
Net charge distribution on each shell is given as
On surface of radius "a"
[tex]q_a = -3.0 mC[/tex]
on radius "b"
[tex]q_b = 3 mC[/tex]
on radius "c"
[tex]q_c = -1.0 mC[/tex]
Now potential at the outer shell is
[tex]V = \frac{kq_c}{r_c}[/tex]
[tex]V = \frac{(9\times 10^9)((-1\times 10^{-3})}{0.09}[/tex]
[tex]V = -1 \times 10^8 Volts[/tex]
Part b)
Since copper sphere is a conducting sphere so here it will be an equi potential surface
So the potential will remain same throughout the surface of this sphere
Now we can say
[tex]V_{inner} = V_{outer} = -1 \times 10^8 volts[/tex]
Part c)
Now electric potential at inner sphere is given as
[tex]V = \frac{kq_a}{r_a} + \frac{kq_b}{r_b} + \frac{kq_c}{r_c}[/tex]
[tex]V = \frac{(9\times 10^9)(-3 mC)}{0.025} + \frac{(9\times 10^9)(3 mC)}{0.06} + \frac{(9\times 10^9)(-1 mC)}{0.09}[/tex]
[tex]V = -7.30 \times 10^8 Volts[/tex]