Sorry for not being in touch for so long....i know exams are coming up for the present students...and they are too busy with their preparations....ALL THE VERY BEST to all of you....
Continuing our trend of mind boggling puzzles....i am shooting you here with a small question...lets see which one among us answers it first....
The Question is:
Put two charged (one positive and the other negative)conducting plates facing each other at a distance of 10 cm. Assume the charge in each plate is 1 Coulomb and each plate has dimensions of 5 cm by 5 cm. If you discharge this capacitor certain amount of energy is released (the spark).
If, before discharging the plates, you pull them apart to reach 15 more centimeters, the energy released when discharged will be smaller or bigger than in the first case? Explain.
Waiting for the answer.........
Take Care....
2 comments:
It will increase, dont ask me why..
Just wanted to be d first one to comment..
Great answer Rahul....its correct...u r the winner for the week....however u did not back up your answer....the correct explanation of the question is:
The energy stored in the electric field of a capacitor is given by
E = (1/2)CV2
The voltage V is proportional to the charge Q, and is given by
V = Q/C
The capacitance, C, is given by
C = KεA/d = Q/V
A is the area of the plates and d the distance of separation between the plates.
From the above equation we see that by increasing the distance between the plates the capacitance decreases. The voltage, however, increases according to the voltage equation, above.
Now, let's re-write the first equation by using the fact that C = Q/V,
E = (1/2)(Q/V)V2 = (1/2)QV
It is clear then - given that V has increased and Q is a constant - that the energy will be bigger.
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