Question

question 20 of 25
a parallel plate capacitor is connected to a battery that has a potential difference of 200 v. once charged, each plate on the capacitor holds a charge of 5.0×10⁻⁹c. what is the capacitance of the parallel plate capacitor? (use c = \\(\frac{q}{\delta v}\\))

a. 5.0×10⁻¹¹f
b. 1.0×10⁻⁶f
c. 2.5×10⁻¹¹f
d. 2.0×10⁻⁶f

Explanation:

Step1: Identify given values

$Q = 5.0\times 10^{-9}\text{C}$, $\Delta V=200\text{V}$

Step2: Apply capacitance formula

$C=\frac{Q}{\Delta V}=\frac{5.0\times 10^{-9}}{200}$
$C = 2.5\times 10^{-11}\text{F}$

Answer:

C. $2.5\times 10^{-11}\text{F}$