When the potential difference between the plates of an ideal air-filled parallel plate capacitor is 35 V, the electric field between the plates has a strength of 750 V/m. If the plate area is 4.0 × 10^-² m², what is the capacitance of this capacitor? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by:

$C =\varepsilon_0\frac{A}{d}$

where $\varepsilon_0$ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

From the problem, we know that the potential difference between the plates is 35 V and the electric field strength between the plates is 750 V/m.

The electric field strength can be related to the potential difference by:

$E = \frac{V}{d}$

Therefore, we can solve for the distance between the plates:

$d = \frac{V}{E} = \frac{35	ext{ V}}{750	ext{ V/m}} = 0.0467	ext{ m}$

Using the given plate area and distance, we can calculate the capacitance:

$C = \varepsilon_0\frac{A}{d} = (8.85	imes 10^{-12}	ext{ F/m})\frac{(4.0	imes 10^{-2}	ext{ m}^2)}{(0.0467	ext{ m})} = 7.6	imes 10^{-12}	ext{ F}$

Therefore, the answer is (B).

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