A parallel-plate capacitor of capacitance 6.0 X10 -10 F is charged to 3.0 V. a.)

Question

A parallel-plate capacitor of capacitance 6.0 X10-10 F is charged to 3.0 V. a.) What is the charge on the positively charged plate? b.) If the spacing between the two plates is 2.0 mm, what isthe electric field in the space between the       plates? A parallel-plate capacitor of capacitance 6.0 X10-10 F is charged to 3.0 V. a.) What is the charge on the positively charged plate? b.) If the spacing between the two plates is 2.0 mm, what isthe electric field in the space between the       plates?

Explanation / Answer

a) Q = C V = 6.0 x 10-10 * 3.0 = 1.8 x 10-9 C =1.8nC b) Q = CV = C E d     ==> E = Q / C d               = 1.8 x 10-9 / 6.0 x 10-10 * 2.0 x10-3               = 0.15 x 104 N / C               = 1.50 kN/C or1.50kV/m

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