An air-tilled parallel-plate capacitor has a capacitance of 4 pF.The plate separ

Question

An air-tilled parallel-plate capacitor has a capacitance of 4 pF.The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 12 pF. The dielectric constant of the wax is: 3.0 6.0 4.0 5.0 A 80 pF capacitor is charged to a potential of 2000 V. How much charge is accumulated on each plate of the capacitor? 1600 C 1250 C 1.6 times 10^-7 C 8.0 times 10^-6 C If the charge on a capacitor is doubled, its stored energy is halved is quartered is doubled is quadrupoled

Explanation / Answer

23) Co = eo*A/d = 4 pF

C = k*eo*A/2d = 12 pF

(k/2)*4 = 12 ===========> k = 6 (Option B)

24) Q = CV = 80*10^-12*2000 = 1.6*10^-7 C (Option C)

25) Energy stored = Q^2/2C   =========> As Q doubles, the energy quadrupoled (option D)

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