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how strong is the electric field between the plates of a 0.80- μf air-gap capacitor if they are 3.0 mm apart and each has a charge of 70 μc ?

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Final answer:

The electric field strength between the plates of a 0.80-μF air-gap capacitor, with each plate having a charge of 70 μC and separated by 3.0 mm, is approximately 2.9 × 10³ V/m.

Step-by-step explanation:

The question concerns calculating the electric field between the plates of a capacitor. The formula to find the electric field (E) between two plates of a capacitor is given by E = V/d, where V is the potential difference and d is the separation between the plates. Since we have only the charge (Q) and the capacitance (C) of the capacitor, we first use the formula Q = CV to find V, where Q is the charge on one plate and C is the capacitance of the capacitor. Once V is found, we plug our values into the electric field formula to find E.

So first, let's find the potential difference (V):

V = Q / C = (70 µC) / (0.80 µF) = 87.5 V

Now, let's find the electric field (E):

E = V / d = 87.5 V / (3.0 mm) = 87.5 V / (0.003 m) = 29167 V/m ≈ 2.9 × 10³ V/m

Therefore, the electric field strength between the plates is approximately 2.9 × 10³ V/m.

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