The two plates of a capacitor hold [tex]+4.5 \times 10^{-3} C[/tex] and [tex]-4.5 \times 10^{-3} C[/tex] of charge when the potential difference is [tex]950 V[/tex]. What is the capacitance?

A. [tex]9.79 \times 10^{-6} F[/tex]
B. [tex]5.44 \times 10^{-7} F[/tex]
C. [tex]3.74 \times 10^{-26} F[/tex]
D. [tex]4.74 \times 10^{-6} F[/tex]



Answer :

To find the capacitance of the capacitor, we can use the formula for capacitance:

[tex]\[ C = \frac{Q}{V} \][/tex]

Here, \( C \) is the capacitance, \( Q \) is the charge stored in the capacitor, and \( V \) is the potential difference applied across the plates.

Given:
- Charge, \( Q = 4.5 \times 10^{-3} \, \text{C} \)
- Potential difference, \( V = 950 \, \text{V} \)

We can substitute these values into the formula:

[tex]\[ C = \frac{4.5 \times 10^{-3}}{950} \][/tex]

When we compute this, we get:

[tex]\[ C = \frac{4.5 \times 10^{-3} \, \text{C}}{950 \, \text{V}} \approx 4.736842105263159 \times 10^{-6} \, \text{F} \][/tex]

It looks like the closest answer to our calculation is:

[tex]\[ 4.74 \times 10^{-6} \, \text{F} \][/tex]

Thus, the correct answer is:

[tex]\[ \boxed{4.74 \times 10^{-6} \, \text{F}} \][/tex]