Question

Two parallel plates 10 cm on a side are given equal and opposite charges of magnitude 5.0 × 10−9 C. The plates are 1.5 mm apart. What is the potential difference between the plates?

Answer 1

`\Delta V = 84.7 volts`

Step-by-step explanation:

As we know that the capacitance of the capacitor is given as

`C = (\epsilon_0 A)/(d)`

now we have

here we know that

Area = (0.10 m)(0.10 m)

distance between plates = 1.5 mm

`C = ((8.85 * 10^(-12))(0.10* 0.10))/(1.5 * 10^(-3))`

now we have

`C = 5.9 * 10^(-11) f`

now potential difference between the plates of capacitor is given as

`\Delta V = (Q)/(C)`

`\Delta V = (5* 10^(-9))/(5.9 * 10^(-11))`

`\Delta V = 84.7 volts`