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Chapter 23: Problem 12

A parallel-plate capacitor is connected to a battery that imposes a potential difference \(V\) between its plates. If a dielectric slab is inserted between the plates, what happens to (a) the potential difference, (b) the capacitor charge, and (c) the capacitance?

Short Answer

Expert verified
When a dielectric slab is inserted into a parallel-plate capacitor, the potential difference \(V\) does not change as it is maintained by the battery, the charge \(Q\) on the capacitor increases to fulfill \(Q = CV\), and the capacitance \(C\) increases due to the presence of the dielectric.

Step by step solution

01

Determine the effect on potential difference

When a dielectric is inserted between the plates of a capacitor that is still connected to the battery, the battery maintains the potential difference \(V\). So, the potential difference does not change.
02

Determine the effect on capacitor charge

Since the potential difference is maintained by the battery (it does not change), and a dielectric increases the capacitance of a capacitor, the charge \(Q\) on the capacitor must increase to fulfill the equation \(Q = CV\).
03

Determine the effect on capacitance

By definition, introduction of a dielectric increases the capacitance of a capacitor. The dielectric reduces the electric field between the plates, allowing the capacitor to store more charge for a given potential difference, thereby increasing the capacitance.

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