Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of $$10^{6}$$ V/m. The plate area is $$10^{-4}$$ m$$^{2}$$. What is the dielectric constant if the capacitance is 15 pF? (given $$\varepsilon_0 = 8.86 \times 10^{-12}$$ C$$^{2}$$/Nm$$^{2}$$)

We are dealing with a parallel-plate capacitor completely filled with a dielectric. For such a capacitor the capacitance is given by the well-known formula

$$C \;=\; \kappa \,\varepsilon_0 \,\dfrac{A}{d},$$

where $$C$$ is the capacitance, $$\kappa$$ is the dielectric constant (relative permittivity) we have to find, $$\varepsilon_0$$ is the permittivity of free space, $$A$$ is the plate area and $$d$$ is the separation between the plates.

The separation $$d$$ is not given directly, but we do know the largest voltage the capacitor can sustain (its voltage rating) and the largest electric field the dielectric can withstand. The electric field in a parallel-plate capacitor is related to the voltage by the basic relation

$$E \;=\; \dfrac{V}{d}.$$

Re-arranging this relation to express $$d$$ in terms of the voltage $$V$$ and the maximum permissible field $$E_{\text{max}}$$, we get

$$d \;=\; \dfrac{V_{\text{max}}}{E_{\text{max}}}.$$

Now we substitute the given numerical values. The maximum voltage is $$V_{\text{max}} = 500 \text{ V}$$ and the maximum field is $$E_{\text{max}} = 10^{6} \text{ V m}^{-1}$$. Hence

$$d \;=\; \dfrac{500}{10^{6}} \text{ m} \;=\; 5 \times 10^{-4} \text{ m}.$$

Next, we substitute this $$d$$ into the capacitance formula to solve for the dielectric constant $$\kappa$$. First rewrite the capacitance formula explicitly for $$\kappa$$:

$$\kappa \;=\; \dfrac{C\,d}{\varepsilon_0\,A}.$$

All quantities in this expression are now known:

Capacitance: $$C = 15 \text{ pF} = 15 \times 10^{-12} \text{ F}$$.

Plate separation: $$d = 5 \times 10^{-4} \text{ m}$$ (found above).

Permittivity of free space: $$\varepsilon_0 = 8.86 \times 10^{-12} \text{ C}^2\text{ N}^{-1}\text{ m}^{-2}$$.

Plate area: $$A = 10^{-4} \text{ m}^2$$.

Substituting each value carefully we obtain

$$\kappa \;=\; \dfrac{(15 \times 10^{-12})\,(5 \times 10^{-4})}{(8.86 \times 10^{-12})\,(10^{-4})}.$$

First, multiply the numbers in the numerator:

$$15 \times 5 = 75,$$

and combine the powers of ten:

$$10^{-12} \times 10^{-4} = 10^{-16}.$$

Thus the numerator becomes

$$75 \times 10^{-16} = 7.5 \times 10^{-15}.$$

For the denominator, multiply the constants:

$$8.86 \times 1 = 8.86,$$

and combine the powers of ten:

$$10^{-12} \times 10^{-4} = 10^{-16}.$$

Hence, the denominator is

$$8.86 \times 10^{-16}.$$

Now we divide numerator by denominator:

$$\kappa \;=\; \dfrac{7.5 \times 10^{-15}}{8.86 \times 10^{-16}}.$$

Because both numerator and denominator contain a factor of $$10^{-16}$$, canceling that common factor leaves a single factor of $$10^{+1}$$ (since $$10^{-15}/10^{-16} = 10^{1}$$):

$$\kappa \;=\; \dfrac{7.5}{8.86}\,\times\,10^{1}.$$

Evaluating the numerical fraction,

$$\dfrac{7.5}{8.86} \approx 0.846.$$

Multiplying by $$10^{1}$$ (that is, by 10) gives

$$\kappa \approx 0.846 \times 10 = 8.46.$$

The value rounds to $$\kappa \approx 8.5$$, matching one of the listed choices.

Hence, the correct answer is Option B.