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3 years ago

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Suppose that we have a 1000pF parallel-plate capacitor with air dielectric charged to 1000V. The capacitor terminals are open ci

rcuited.
Find the stored energy. If the plates are moved farther apart so that d is doubled, determine the new voltage on the capacitor and the new stored energy.

Where did the extra energy come from?

Please show all workings & calculations on how you came to your answer.

1 answer:

N76 [4]3 years ago

4 0

Answer:

W = 1 mJ

V_new = 2000 V

W_new = 2 mJ

The extra energy came from the work done from moving the plates

Explanation:

We are given;

Capacitance; C = 1000pF = 10^(-9) F

Voltage; V = 1000V

Now,formula for stored energy in a parallel plate capacitor is given by;

W = (1/2)CV² = (1/2)(10^(-9))(1000²) = 0.5 mJ

However, in this case, it's W = 0.5 x 2 = 1 mJ since parallel-plate capacitor with air dielectric

When plates are moved and distance between plates is doubled(net charge is the same), thus we can calculate voltage from;

Q = CV

Since, C_new = C/2

Thus,

Q = (C/2)V_new

V_new = 2Q/C

Thus, V_new = 2V

Thus, V_new = 2 x 1000 = 2000 V

Now,

W_new = (1/2)C_new•(V_new)² = (1/2) (0.5C)•(2V)² = CV² = 2W = 2 x 1 = 2mJ

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$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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4 years ago