Question

A parallel-plate vacuum capacitor has 6.22 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.60 mm, What is the maximum magnitude of charge that can be placed on each plate if the electric tiekd in the region between the plates is not to exceed 3.00×10⁴ V/in? A parallel-plate capacitor has capacitance C =8.20pF when there is air between the plates. The separation between the Express your answer with the appropriate units.

Answer 1

The maximum magnitude of charge that can be placed on each plate of a parallel-plate vacuum capacitor can be calculated using the formula for electric field between the plates and the formula for energy stored in a capacitor.

Step-by-step explanation:

The maximum magnitude of charge that can be placed on each plate of a parallel-plate vacuum capacitor can be calculated using the formula for electric field between the plates and the formula for energy stored in a capacitor.

First, we need to find the electric field between the plates. We can use the formula:

E = V / d

Where E is the electric field, V is the voltage, and d is the separation between the plates. Substituting the given values, we have:

E = (3.00×10⁴ V/in) * (1 in / 25.4 mm) * (1.60 mm) = 2.99 × 10³ V/m

Next, we can calculate the capacitance of the capacitor using the formula:

C = Q / V

Where C is the capacitance, Q is the charge on each plate, and V is the voltage. Rearranging the formula, we have:

Q = C * V

Substituting the given values, we have:

Q = (8.20 pF) * (3.00×10³ V) = 2.46 × 10⁻² µC

Therefore, the maximum magnitude of charge that can be placed on each plate is 2.46 × 10⁻² µC.