A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is $$7 \times 10^{8}\,\text{V/s}$$ then the integer value of the distance between the parallel plates is (Take $$\varepsilon_0 = 9 \times 10^{-12}\,\frac{F}{m},\ \pi = \frac{22}{7}$$), ________ $$\mu\text{m}$$.

For a parallel plate capacitor, the capacitance is given by $$C = \frac{\varepsilon_0 A}{d}$$.

The charging current satisfies $$I = C\frac{dV}{dt}$$, so $$C = \frac{I}{dV/dt}$$. Equating this with the capacitance expression yields

$$d = \frac{\varepsilon_0 A}{C} = \frac{\varepsilon_0 A \cdot (dV/dt)}{I}$$.

Here, $$\varepsilon_0 = 9 \times 10^{-12}\text{ F/m}$$, the plate radius is $$r = 10\text{ cm} = 0.1\text{ m}$$ so $$A = \pi r^2 = \pi(0.01)$$, the current is $$I = 0.15\text{ A}$$, and $$\frac{dV}{dt} = 7 \times 10^8\text{ V/s}\,$$.

Substituting these values into the expression for \(d\) gives

$$d = \frac{9 \times 10^{-12} \times \frac{22}{7} \times 0.01 \times 7 \times 10^8}{0.15}$$

which simplifies to

$$= \frac{9 \times 10^{-12} \times 22 \times 0.01 \times 10^8}{0.15}$$

$$= \frac{9 \times 22 \times 10^{-12+8-2}}{0.15} = \frac{198 \times 10^{-6}}{0.15}$$

$$= 1320 \times 10^{-6} \text{ m} = 1320 \,\mu\text{m}$$

The answer is 1320 μm.