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Chapter 13: Problem 29

A capacitor is connected to a battery. The force of attraction between the plates when the separation between them is halved (A) remains the same. (B) becomes eight times. (C) becomes four times. (D) becomes double.

Short Answer

Expert verified
When the separation between the plates of a capacitor is halved, the force of attraction between the plates becomes double. Answer: (D) becomes double.

Step by step solution

01

Recall the formula for the force of attraction

The force of attraction (F) between the plates of a capacitor with a charge Q and area A separated by a distance d can be expressed as follows: \[F = \frac{Q^2}{2\epsilon_0 A}\] Where 蔚鈧� is the vacuum permittivity constant, given by: \[\epsilon_0 = 8.85 \times 10^{-12} \frac{C^2}{N.m^2}\] The capacitance (C) of the capacitor is given by: \[C = \epsilon_0\frac{A}{d}\] Since the capacitor is connected to a battery, it maintains a constant charge (Q) on the plates.
02

Find the expression for the force when the separation is halved

Now we want to find the force of attraction between the plates when the separation between them is halved. Let d' be the new separation which is given by \(d' = \frac{d}{2}\). The new capacitance, C', becomes: \[C' = \epsilon_0\frac{A}{d'}\] We can express the force of attraction in terms of the new separation, d'. \[F' = \frac{Q^2}{2\epsilon_0 A'}\] Since the area remains the same and charge Q is constant, we can compare the two expressions for the force of attraction: F' versus F.
03

Determine the change in the force of attraction

By dividing F' by F, we can determine the change in the force of attraction when the separation is halved: \[\frac{F'}{F} = \frac{\frac{Q^2}{2\epsilon_0 A'}}{\frac{Q^2}{2\epsilon_0 A}}\] Simplifying, we get: \[\frac{F'}{F} = \frac{A'}{A}\] Now substitute the expressions for A' and A in terms of the capacitor and original plate separation: \[\frac{F'}{F} = \frac{\epsilon_0 \frac{A}{d'}}{\epsilon_0\frac{A}{d}} = \frac{d}{d'}\] Since \(d' = \frac{d}{2}\), we have: \[\frac{F'}{F} = \frac{d}{\frac{d}{2}} = 2\]
04

Conclusion

So, when the separation between the plates of the capacitor is halved, the force of attraction between the plates becomes double. Therefore, the correct answer is: (D) becomes double.

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