The potential inside the inner sphere will be "0".
The inside sphere would be charged positively. Increased potential expressions ([tex]V_1[/tex]) because of the interior metal sphere, such as:
- [tex]V_1 = \frac{kq}{r}[/tex]
The outer area sphere would be charged positively. Increased potential expressions ([tex]V_2[/tex]) because of the interior metal sphere, such as:
- [tex]V_2 = \frac{k(-q)}{r}[/tex]
hence,
Throughout the inner sphere, the potential will be:
→ [tex]V = V_1+V_2[/tex]
By substituting the values, we get
→ [tex]= \frac{kq}{r} +(-\frac{kq}{r} )[/tex]
→ [tex]= \frac{kq}{r} -\frac{kq}{r}[/tex]
→ [tex]= 0[/tex]
Thus the answer above is right.
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