Question

A parallel-plate capacitor consists of two identical, parallel, conducting plates each with an area of 5.0 cm2 and uniform charges of +8.4 pC or -8.4 pC. The plates are separated by a perpendicular distance of 1.0 mm. What is the potential difference across the metallic plates?

Answer 1

Given:

Area = 5.0 cm²

Uniform charges = +8.4 pC or -8.4 pC

Perendicular distance = 1.0 mm

Let's find the potential difference across the metallic plates.

To find the potential difference, apply the formula:

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

Where:

V is the potential difference

Q is the charge = 8.4 pC

C is the capacitance.

To find C, apply the formula:

$\begin{gathered} C=\frac{E_oA}{D} \\ \end{gathered}$

Where:

Eo = 8.85 x 10⁻¹² f/m

A is the area

D is the distance.

Thus, we have:

$\begin{gathered} C=\frac{(8.85\times10^{-12})(5.0\times10^{-4})}{1.0\times10^{-3}} \\ \\ C=4.425\times10^{-12}\text{ F} \end{gathered}$

Now, to find the potential difference, we have:

$\begin{gathered} V=\frac{Q}{C} \\ \\ V=\frac{8.4\times10^{-12}}{4.425\times10^{-12}} \\ \\ V=1.89\text{ V }\approx1.9\text{ V} \end{gathered}$

The potential difference across the metallic plates is 1.9 volts

ANSWER:

$1.9\text{ V}$