Question

2. A parallel-plate capacitor has an area of 4.0 cm2, and the plates are separated by 1.0 mm.
a. What is the capacitance?
b. How much charge does this capacitor store when connected to a 9.0 V battery?

Answer 1

Explanation:

$Area \: of \: plates= 4.0 \: {cm}^{2} = 4.0 \times {10}^{ - 4} \: {m}^{2} \\ distance \: between \: plates = 1.0 \: mm = 1.0 \times {10}^{ - 3} m \\ capacitance = \epsilon_{0} \times \frac{area \: of \: plate}{distance \: between \: plates} \\ wher \: \epsilon_{0} = 8.84 \times {10}^{ - 12} \\ capacitance = 8.84 \times {10}^{ - 12} \times \frac{4 \times {10}^{ - 4} }{1.0 \times {10}^{ - 3} } \\ = 8.84 \times 4 \times {10}^{ - 12 - 4 + 3} farad \\ = 35.36 \times {10}^{ - 13} farad \\ = 3.536 \times {10}^{ - 12} farad \\ = 3.5 \: p f$

$capacitance \: = \: 3.5 \: pf \\ v = 9.0 \: volt \\ charge \: q = \: c \times v = 3.5 \times 9 = 31.5 \: pc$

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