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

Answer:

$\Delta V = 84.7 volts$

Explanation:

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

$C = \frac{\epsilon_0 A}{d}$

now we have

here we know that

Area = (0.10 m)(0.10 m)

distance between plates = 1.5 mm

$C = \frac{(8.85 \times 10^{-12})(0.10\times 0.10)}{1.5 \times 10^{-3}}$

now we have

$C = 5.9 \times 10^{-11} f$

now potential difference between the plates of capacitor is given as

$\Delta V = \frac{Q}{C}$

$\Delta V = \frac{5\times 10^{-9}}{5.9 \times 10^{-11}}$

$\Delta V = 84.7 volts$