4:10 PM Sprint 17% webassign net My Notes o Ask Your Teacher Question Details Se

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4:10 PM Sprint 17% webassign net My Notes o Ask Your Teacher Question Details SerCPB 16 P004.son. An ion accelerated through 300.00 v has its potential energy decreased by potential energy of 9.60x 1017 J, Calculate the magnitude of the) charge on the ion. C Need Help? Show My Work Question Details SerCPB 16-P027. My Notes An air-filled parallel-plate capacitor has plates of area 2.70 cm2 separated by o.so mm. The capacitor is connected to a 15.0 v battery. (a) Find the value of its (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates? Need Help? Read LTaate Tute optional? Show My Work My Notes C Ask Your Teache Question Details SerCPB 16 P003 ssm potential difference of 75 mv exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to edect a positive sodium ion (Na from the interior of the cell? Need Help? L ReadI L O Show My Work optional? Save Assignment Extension Requesr Home My Assignments

Explanation / Answer

Q8.

change in potential energy=charge*potential difference

==>9.6*10^(-17)=charge*300

==>charge=3.2*10^(-19) C

Q9.

part a:

capacitance=epsilon*area/distance

=8.85*10^(-12)*2.7*10^(-4)/(0.5*10^(-3))

=4.779 pF

part b:

charge on capacitor=capacitance*voltage

=71.685 pC

part c:

magnitude of electric field=potential difference/distance

=15/(0.5*10^(-3))=3*10^4 N/C

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