a) A parallel-platecapacitor of dimensions 2.15 cm × 5.69 cm is separated by a1.
ID: 1672933 • Letter: A
Question
a) A parallel-platecapacitor of dimensions 2.15 cm × 5.69 cm is separated by a1.65 mm thickness of paper. Find the capacitance of this device.The dielectric constant for paper is 3.7.Answer in units of pF.
b) What is the maximum charge that can be placed on the capacitor?The electric strength of paper is 1.6 × 107 V/m.
Answer in units of c.
a) A parallel-platecapacitor of dimensions 2.15 cm × 5.69 cm is separated by a1.65 mm thickness of paper. Find the capacitance of this device.The dielectric constant for paper is 3.7.
Answer in units of pF.
b) What is the maximum charge that can be placed on the capacitor?The electric strength of paper is 1.6 × 107 V/m.
Answer in units of c.
Explanation / Answer
Area A = 2.15 cm * 5.69 cm = 12.2335 cm ^ 2 = 12.2335 * 10^ -4 m^ 2
Thickness d = 1.65 mm= 1.65 * 10 ^ -3 m
Dielectric constant of paper k = 3.7
Capacitance C = k A / d
Where = 8.85 * 10 ^ -12 c ^ 2/ N m^2
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