answer correctly and ill rate you 5 stars A parallel-plate capacitor in air has

Question

answer correctly and ill rate you 5 stars

A parallel-plate capacitor in air has a plate separation of 1.29 cm and a plate area of 25.0 cm2 . The plates are charged to a potential difference of 240 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine the charge on the plates before and after immersion, before pC after pC Determine the capacitance and potential difference after immersion. Cf = F Delta Vf = V Determine the change in energy of the capacitor. nJ

Explanation / Answer

a) C = er .e0 .A/d   = 8.854 x 10-12 x 1 x 25 x 10-4 / 0.0129

       = 1.72 x 10-12 F

Q = CV = 240 x 1.72 x 10-12 = 4.12 x 10-10 C = 0.412 pC   .....before

after will be same = 412 pC

b)

after C = er .e0 .A/d   = 8.854 x 10-12 x78.52 x 25 x 10-4 / 0.0129
       = 135.05 x 10-12 F = 135.05 pC

V = Q/C = 0.412 x 10-9 / 135.05x10-12   =3.05 V

c) change in energy = 135.05 x 10-12 x 3.05^2 /2    -   0.412x10-12 x 240^2 /2  

= -1.124 x 10-8 J = - 11.24x 10-9 J = - 11.24 nJ

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