In: Physics
Three capacitors, all initially without any dielectrics inserted, of C1 = 4 micro-farads, C2 = 10 micro-farads, and C3 = 10 micro-farads, are connected purely in series to a battery and the total stored energy of all three capacitors is 248 micro-joules. Then, the plates of C1 are brought closer by a factor of 3 and a dielectric with a dielectric constant of 1.9 is inserted into C2 while still connected in series to the same battery. (C3 does not change at all.) How much does the total stored energy of all three capacitors increase in micro-joules after the capacitors are changed? Not the factor change... how much does the stored energy go up in value. step by step please and thank you!
C1 = 4 F
C2 = 10 F
C3 = 10 F
Series combination of the capacitors is given as
1/Cs = 1/C1 + 1/C2+ 1/C3
1/Cs = 1/4 + 1/10+ 1/10
Cs = 20/9 F
V = Battery Voltage = ?
U = Total energy stored in the series combination of the capacitors = 248 J
The total energy stored in the series combination of the capacitors is given as
U = (0.5) Cs V2
248 = (0.5) (20/9) V2
V = 14.94 Volts
we know that
C = A/d where A = area of the plate, d = distance between the plates
the capacitance is inversely related to the distance between the plates.
Hence, as the plates of capacitor C1 are brought closer by a factor of 3, the capacitance increases by a factor of 3.
C1' = capacitance of capacitor C1 after plates are brought closer = 3 C1 = 3 x 4 = 12 F
k = dielectic constant of the dielectric inserted between the plates of capacitor C2 = 1.9
C2' = capacitance of capacitor C2 after dielectric is inserted = k C2 = 1.9 x 10 = 19 F
series combination of capacitors is now given as
1/C's = 1/C'1 + 1/C'2+ 1/C3
1/C's = 1/12 + 1/19+ 1/10
C's = 4.24 F
The total energy stored in the new series combination of the capacitors is given as
U' = (0.5) C's V2
U' = (0.5) (4.24) (14.94)2
U' = 473.2 J
Increase in energy is given as
U = U' - U = 473.2 - 248 = 225.2 J