A parallel plate capacitor has a capacitance of 1.71 nF. There is a charge of 0.809 μC on each plate.

To find the voltage across the plates of a parallel plate capacitor, you can use the formula:

V = Q / C

where V is the voltage, Q is the charge on each plate, and C is the capacitance.

In this case, the charge on each plate (Q) is given as 0.809 μC (microcoulombs) and the capacitance (C) is given as 1.71 nF (nanofarads).

First, let's convert the charge from microcoulombs to coulombs:

0.809 μC = 0.809 × 10^(-6) C

Next, let's convert the capacitance from nanofarads to farads:

1.71 nF = 1.71 × 10^(-9) F

Now, we can substitute these values into the formula to find the voltage (V):

V = (0.809 × 10^(-6) C) / (1.71 × 10^(-9) F)

We can simplify this expression by canceling out the common factors:

V = (0.809 × 10^(-6)) / (1.71 × 10^(-9))

V = 0.809 / 1.71 × 10^(-9 -(-6))

V = 0.809 / 1.71 × 10^(-9 + 6)

V = 0.809 / 1.71 × 10^(-3)

V = 0.473 V

Therefore, the voltage across the plates of the parallel plate capacitor is 0.473 V.