How much charge can be placed on a capacitor of plate area 15 cm² with air between the plates before it reaches "atmospheric breakdown" where E = 3.0 × 10⁶ V/m? (ε₀ = 8.85 × 10−¹² C²/N⋅m²) A)4.0E-8 C B)1.3E-8 C C)1.8E-10 C D)5.1E-8 C E)1.7E-8 C

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Quizplus · Accepted Answer

The electric field between the plates of a capacitor is given by:$$E = \frac{V}{d}$$where:* E is the electric field in V/m* V is the voltage across the capacitor in V* d is the distance between the plates in mThe maximum electric field that can be sustained in air before it breaks down is called the "atmospheric breakdown" field strength. This value is approximately 3.0 × 10^6 V/m.In this problem, we are given the plate area (A = 15 cm^2 = 1.5 × 10^-3 m^2), the distance between the plates (d = 0.1 cm = 1.0 × 10^-3 m), and the atmospheric breakdown field strength (E = 3.0 × 10^6 V/m). We can use these values to calculate the maximum charge that can be placed on the capacitor before it reaches atmospheric breakdown.The capacitance of a capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$where:* C is the capacitance in F* ε0 is the permittivity of free space (8.85 × 10^-12 C^2/N⋅m^2)* A is the plate area in m^2* d is the distance between the plates in mSubstituting the given values into this equation, we get:$$C = \frac{(8.85 	imes 10^{-12} 	ext{ C}^2/	ext{N}\cdot	ext{m}^2)(1.5 	imes 10^{-3} 	ext{ m}^2)}{1.0 	imes 10^{-3} 	ext{ m}} = 1.33 	imes 10^{-11} 	ext{ F}$$The charge on a capacitor is given by:$$Q = CV$$where:* Q is the charge in C* C is the capacitance in F* V is the voltage across the capacitor in VSubstituting the given values into this equation, we get:$$Q = (1.33 	imes 10^{-11} 	ext{ F})(3.0 	imes 10^6 	ext{ V}) = 4.0 	imes 10^{-5} 	ext{ C}$$Therefore, the maximum charge that can be placed on the capacitor before it reaches atmospheric breakdown is 4.0 × 10^-5 C.

How Much Charge Can Be Placed on a Capacitor of Plate