Question

wo plates with area 6.00×10−3 m2 are separated by a distance of 1.50×10−4 m . If a charge of 5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.

Answer 1

Answer:

V = 152.542 volts

Explanation:

Given data:

area of plates$6.00\times 10^{-3} m^2$

distance between the plates is $1.50\times 10^{-4} m$

charge = $5.40\times 10^{-8} c$

we know that capacitance is given as

$C = \frac{\epsilon A}{d}$

$C = \frac{8.85\times 10^{-12} 6\times 10^{-3}}{1.50\times 10^{-4}}$

$C = 3.54\times 10^{-10} F$

potential difference is given as

$V =\frac{Q}{C} = \frac{5.40\times 10^{-8}}{3.54\times 10^{-10}}$

V = 152.542 volts