Animated Solution for Physics - Electrostatics: Voltage rating of a parallel plate capacitor is $500\text{ V}$. Its dielectric can withstand a maximum electric field of $10^6\text{ V/m}$. The plate area is $10^{-4}{\text{ m}}^2$. What is the dielectric constant, if the capacitance is $15\text{ pF}$? (Take, ${\epsilon }_0=8.86\times 10^{-12}{\text{ C}}^2/{\text{N-m}}^2$)

Imagine you are designing a circuit and you need a capacitor that can store a specific amount of charge while withstanding a high voltage. The problem presents us with a parallel plate capacitor filled with a dielectric material. We are given its vital statistics: a voltage rating of $500\text{ V}$, a maximum electric field tolerance of $10^6\text{ V/m}$, a plate area of $10^{-4}{\text{ m}}^2$, and a capacitance of $15\text{ pF}$.

Our mission is to uncover the identity of the dielectric material by calculating its dielectric constant, $K$.

Voltage, Field, and Distance

To find $K$, we naturally turn to the fundamental formula for the capacitance of a parallel plate capacitor with a dielectric:

We know $C$, ${\epsilon }_0$, and $A$. But there is a missing piece in our puzzle: the distance $d$ between the plates. How do we find it?

This is where the voltage rating and the maximum electric field come into play. The maximum electric field a dielectric can withstand before it breaks down and starts conducting is called its dielectric strength. In a uniform electric field, the relationship between voltage $V$, electric field $E$, and distance $d$ is beautifully simple:

By rearranging this, we can express the unknown distance $d$ in terms of the known voltage and electric field:

Now, we can substitute this expression for $d$ back into our capacitance formula. This elegant substitution eliminates the unknown variable and gives us an equation entirely in terms of known quantities:

Since we want to find the dielectric constant $K$, we rearrange the equation to isolate it:

With our master equation ready, it is time to plug in the values. This is where we must be extremely careful with units. The capacitance is given in pico-farads ($\text{pF}$), which must be converted to farads ($\text{F}$) by multiplying by $10^{-12}$.

Let's simplify the numerator and the denominator separately to avoid errors.

The numerator becomes:

The denominator becomes:

Now, we divide the two:

Performing the final division yields $K\approx 8.465$. Rounding this to one decimal place to match our options, we arrive at our final answer:

This tells us that the dielectric material inside the capacitor increases its capacitance by a factor of 8.5 compared to a vacuum.