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physics 2 14 etisalat 4G 11:01 PM e edugen.wileyplus.com WileyPLUS ASSIOMENT SES

ID: 1628877 • Letter: P

Question


physics 2 14
etisalat 4G 11:01 PM e edugen.wileyplus.com WileyPLUS ASSIOMENT SESOUChapter 25, W2 2Chapter 25, Problem 012 ce23 V bttery. Two parallel-plate capacitors, 6.3 F each, are connected in parallel to a 23 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional 2.charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors? Cick if you weuls ike to Shew Work for this questionc 0 War Review Scene

Explanation / Answer

plate distance is halved means capacitance is doubled

new capacitance of squeezed capacitor = 2 * 6.3 = 12.6F

new total capacitance = 12.6F + 6.3F = 18.9F

old total capacitance = 6.3F + 6.3F = 12..6F

difference = 18.9 - 12.6 = 6.3F

(a)Q = C * V = 6.3 F * 23 V = 144.9 x 10^-6 = 1.449 x 10^-4 C

(b) The increase in the total charge on both capacitors is 1.449 x 10^-4 C (the additional charge was transferred to the SQUEEZED CAPACITOR ONLY)