Find Dielectric Constant from Parallel Plate Capacitor Problem

AI Thread Summary
To find the dielectric constant of a parallel plate capacitor with a capacitance of 1.5µF and a potential difference of 5.0V after inserting a dielectric, the charge remains constant since the battery is removed. The initial voltage across the capacitor was 12V, and the capacitance increases with the dielectric. The relevant equation is Q = CV, where Q is the charge, C is capacitance, and V is voltage. By calculating the new capacitance with the dielectric and comparing it to the original capacitance, the dielectric constant can be determined. This approach effectively utilizes the relationship between charge, capacitance, and voltage to solve for the dielectric constant.
JordanHood
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Homework Statement


A parallel plate capacitor has a capacitance of 1.5µF with air between the plates. The capacitor is connected a 12V battery and charged. The battery is then removed. When a dielectric is placed between the plates, a potential difference of 5.0V is measured across the plates. What is the dielectric constant of the material?

Homework Equations

The Attempt at a Solution


I know that Q=CV. however this is the only equation I find relevant as all the other equations I know for finding the dielectric constant involve area and distance between the plates. Can someone just help me get started with this question?
 
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Inserting a dielectric will increase the capacitance but charge on the the plates will be same. Can you write an equation with this information and find the new capacitance? Comparing the two capacitances, you can calculate the dielectric constant of the new capacitor.
 
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