Five identical 10.0 mu F capacitors are arranged as shown. The capacitors are co

Question

Five identical 10.0 mu F capacitors are arranged as shown. The capacitors are connected to a 12.0 V battery. What is the charge on C1 when the capacitors are fully charged? You have an air-filled capacitor composed of square parallel plates (each plate 3.00 cm per side) When connected to a 12.0 V battery, the capacitor can store 1.40 Times 10-11 J of energy when fully charged. What is the separation between the plates of the capacitor? A certain dipole consists of two charges (+2e and -2e) separated by a distance of 1.50 nm. It is placed between the plates of the fully-charged capacitor from part (b). What is the maximum potential energy of the dipole?

Explanation / Answer

a) as you have calculated, Ceq=5 uF

then net charge=12*5=60 uC

now C5 will have charge of 60 uC.(as in series connection, every capacitor has same charge)

then voltage across C5=charge/capacitance=60 uC/10 uF=6 volts

then voltage across the parallel combination =12-6=6 volts

hence voltage across combination of C1 and C2 is 6 volt.

then as both have same capacitance, voltage will be divided equally

hence voltage across C1=6/2=3 volts

then charge on C1=3*10=30 uC

b)

energy=0.5*capacitance*voltage^2

==>1.4*10^(-11)=0.5*C*12^2

==>C=1.94*10^(-13) F

as we know, capacitance of parallel plate capacitor=epsilon*area/distance

==>distance=epsilon*area/capacitance

=8.85*10^(-12)*0.03*0.03/(1.94*10^(-13))=0.0411 m=4.11 cm

c)dipole moment=charge*distance=2*e*1.5*10^(-9)

maximum potential energy=electric field*dipole moment

=(voltage/distance)*dipole moment

=(12/0.0411)*2*1.6*10^(-19)*1.5*10^(-9)
=1.4*10^(-25) J

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