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WlleyPLUS: MyWileyPLUS I Help I Contact Us I Log Cutnell, Physics, 9e College Physics I and II (PHY 2053/205 Assignment Gradebook ent FULL SCREEN PRINTER VERSION BACK NEXT Chapter 19, Problem 54 An empty parallel plate capacitor is connected between the terminals of a 11.7-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor? Number the tolerance is +/-296 Click if you would like to Show Work for this question: Open Show Work Units Question Attempts: 0 of 5 used SAVE POR LATER. SUBMIT ANSWER Copyright© 2000-2015 by John Wiley & Sons, Inc. or related companies. All rights reserved.

Explanation / Answer

capacitor of a parallel plate capacitor is given as

C=epsilon*area/distance between plates

where epsilon=electrical permitivity of the medium

after being charged by voltage V, charge storage in the capacitor=capacitance*voltage=C*V

once the capacitor is disconnected, the charge remains constant (assuming capacitor is ideal)

then as spacing between plates is doubled, then capacitor will become half.

then as charge remains constant,

C*V=constant

==> C*11.7=(C/2)*V

==> V=23.4 volts

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