Answer:
(a) Capacitance is [tex]1\mu C[/tex]
(b) Potential difference is 256 V
Solution:
As per the question:
Charges on the two conductors, q is [tex]+ 16\muC[/tex] and [tex]- 16\muC[/tex]
The potential difference between the charges, V = 16.0 V
Now,
(a) The capacitance of the system is given by:
q = CV
[tex]C = \frac{q}{V} = \frac{16\times 10^{- 6}}{16} = 1\mu F[/tex]
(Since, [tex]1\mu C = 10^{- 6} C[/tex])
Now,
(b) When the charge, 'q' is increased to [tex]+ 256\mu C[/tex] and [tex]- 256\mu C[/tex]
q = CV
[tex]V = \frac{q}{C} = \frac{256\times 10^{- 6}}{1\times 10^{- 6}} = 256\ V[/tex]
