Answer: C.
Explanation:
For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.
Now, the stored energy can be written as:
E = (1/2)*Q^2/C
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.
If the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.
The given problem is based on the concept of parallel plat capacitor. For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
here.
e is the permittivity of free space.
Since, the distance is inversely proportional then if we double the distance, the capacitance halves. Â Now, the stored energy can be given as,
E = (1/2)*Q^2/C
here,
Q is the charge stored in the capacitor.
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
So, the energy is proportional to the distance between the plates.
Thus, we can conclude that if the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.
Learn more about the energy stored in a capacitor here:
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