Answer:
Part a)
tex]V = -1 \times 10^8 Volts[/tex]
Part b)
![V_(inner) = V_(outer) = -1 * 10^8 volts](https://img.qammunity.org/2020/formulas/physics/college/ttbenr8ro7wt5y04v9s9cg2bmg0y20ld6b.png)
Part c)
![V = -7.30 * 10^8 Volts](https://img.qammunity.org/2020/formulas/physics/college/ajb60ibvtk5zzfhcc4ncwk7d913kty77h7.png)
Step-by-step explanation:
Part a)
Net charge distribution on each shell is given as
On surface of radius "a"
![q_a = -3.0 mC](https://img.qammunity.org/2020/formulas/physics/college/5xv57e8cxa6n1ziwjwfknh2h2q2k8t3bay.png)
on radius "b"
![q_b = 3 mC](https://img.qammunity.org/2020/formulas/physics/college/k5ay4dyo388f2ji0l47m48dmqpsvy3z4bz.png)
on radius "c"
![q_c = -1.0 mC](https://img.qammunity.org/2020/formulas/physics/college/hhqj4jx1y0fy1lrtpexnxlb8selvede7cr.png)
Now potential at the outer shell is
![V = (kq_c)/(r_c)](https://img.qammunity.org/2020/formulas/physics/college/fhqsughhoeh9pg19727n4e0r79uuj39fpg.png)
![V = ((9* 10^9)((-1* 10^(-3)))/(0.09)](https://img.qammunity.org/2020/formulas/physics/college/sytbabjbghkliho01d06mmgzims18x7is6.png)
![V = -1 * 10^8 Volts](https://img.qammunity.org/2020/formulas/physics/college/ow4dcafsvuh9czwvvof1u3pzob9jmdm2ax.png)
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
![V_(inner) = V_(outer) = -1 * 10^8 volts](https://img.qammunity.org/2020/formulas/physics/college/ttbenr8ro7wt5y04v9s9cg2bmg0y20ld6b.png)
Part c)
Now electric potential at inner sphere is given as
![V = (kq_a)/(r_a) + (kq_b)/(r_b) + (kq_c)/(r_c)](https://img.qammunity.org/2020/formulas/physics/college/f2jo1mosrt0o7acyu9tm8zld6uaj4h1ol2.png)
![V = ((9* 10^9)(-3 mC))/(0.025) + ((9* 10^9)(3 mC))/(0.06) + ((9* 10^9)(-1 mC))/(0.09)](https://img.qammunity.org/2020/formulas/physics/college/qjt59z4n87h618i0brspk0heldapm5jp7z.png)
![V = -7.30 * 10^8 Volts](https://img.qammunity.org/2020/formulas/physics/college/ajb60ibvtk5zzfhcc4ncwk7d913kty77h7.png)