. A 10 micro-coulomb capacitor wasconstructed with a parallel plate geometry. A

Question

. A 10 micro-coulomb capacitor wasconstructed with a parallel plate geometry. A 400 sq cm metalplate was separated from another similar plate by 2 cm. Whatwould be the work done in Joules when the plate separation wasmanufactured as 5 cm?

Area (A) of each square plate of the capacitor = 400cm2

                                                                      =4.0*10-2m2

seperation (d) between the plates = 5 cm = 0.05m

Now capacitance (C) of the capacitor is

              C = Ao /d              

Then workdone is U = Q2 / 2C

Here Q = 10.0*10-6C

Do the calculation work!

Whats the answer?

Explanation / Answer

Area (A) of each square plate of the capacitor = 400cm2

                                                                      =4.0*10-2m2

seperation (d) between the plates = 5 cm = 0.05m

Now capacitance (C) of the capacitor is

              C = Ao /d              

                = 708 * 10 ^ -14 F

Then workdone is W = Q2 / 2C

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.