The potential difference required is 9.0 V
Explanation:
The relationship between the capacitance, the potential difference and the charge stored in a capacitor is the following:
[tex]C=\frac{Q}{V}[/tex]
where
Q is the charge stored
V is the potential difference
C is the capacitance
For the capacitor in this problem, we have
[tex]C=2.00 pF = 2.0\cdot 10^{-12} F[/tex] is the capacitance
[tex]Q=18.0 pF = 18.0\cdot 10^{-12} C[/tex] is the charge stored
Solving the equation for V, we find the potential difference on the capacitor:
[tex]V=\frac{Q}{C}=\frac{18\cdot 10^{-12}}{2.0 \cdot 10^{-12}}=9.0 V[/tex]
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The potential difference will required to 9 Volt.
The potential require by capacitor is given as,
[tex]V=\frac{Q}{C}[/tex]
Where Q is charge and C is capacitance.
Given that, [tex]C=2pF=2*10^{-12}F[/tex] and [tex]Q=18pC=18*10^{-12}C[/tex]
Substitute values in above equation.
[tex]V=\frac{18*10^{-12} }{2*10^{-12}} \\\\V=9Volt[/tex]
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