Answer:
V = 152.542 volts
Explanation:
Given data:
area of plates[tex] 6.00\times 10^{-3} m^2[/tex]
distance between the plates is [tex]1.50\times 10^{-4} m[/tex]
charge = [tex]5.40\times 10^{-8} c[/tex]
we know that capacitance is given as
[tex] C = \frac{\epsilon A}{d}[/tex]
[tex]C = \frac{8.85\times 10^{-12} 6\times 10^{-3}}{1.50\times 10^{-4}}[/tex]
[tex]C = 3.54\times 10^{-10} F [/tex]
potential difference is given as
[tex]V =\frac{Q}{C} = \frac{5.40\times 10^{-8}}{3.54\times 10^{-10}}[/tex]
V = 152.542 volts
