A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950MNaOH. The acid req
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Answer:
A) increase.
Explanation:
- By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:
- At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:
C = ε₀*A / d (2)
- If the space between plates, is filled with a dielectric of dielectric constant κ, the above equation becomes:
- If the capacitor, once charged, is disconnected from the battery, charge must keep the same.
- Now, if we remove the dielectric, as stated in (3) and (2), the capacitance C will decrease when removing the dielectric.
- From (1) if C decreases, and Q remains constant, in order to keep both sides of the equation equal each other, V (the potential difference between plates), must increase.