Question

When a 360 nF air capacitor is connected to a power supply, the energy stored in the capacitor is 1.85 x 10-5 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32 x 10-5 J. (a) What is the potential difference between the capacitor plates

Answer 1

Answer:

(a) Approximately $10.1\; {\rm V}$.

Explanation:

Let $C$ denote the capacitance of a capacitor. Let $V$ be the potential difference (voltage) between the two plates of this capacitor. The energy $E$ stored in this capacitor would be:

$\displaystyle E = \frac{1}{2}\, C\, (V^{2})$.

Rearrange this equation to find an expression for the potential difference $V$ in terms of capacitance $C$ and energy $E$:

$\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}$.

$\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}$

The capacitance $C$ of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

$\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}$.

Given that the energy stored in this capacitor is $E = 1.85 \times 10^{-5}\; {\rm J}$, the potential difference across the capacitor plates would be:

$\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}$.