Question

A parallel-plate capacitor made of half-circular plates of radius 10 cm separated by 1.50 mm is charged to a potential difference of 600 volts by a battery. Then a sheet of tantalum pentoxide is pushed between the plates, completely filling the gap between them. How much additional charge flows from the battery to one of the plates when the tantalum pentoxide is inserted? Dielectric constant for tantalum pentoxide: 25

Answer 1

capacitance of capacitor before sheet is inserted,

C = e0*A / d

C = 8.85*10^(-12)*pi*(0.10)^2 / 1.5*10^(-3)

C = 1.85*10^(-10) F

Change on capacitor,

Q = C*V

Q = 1.85*10^(-10) * 600 = 1.112*10^(-7) C

When  sheet of tantalum pentoxide is pushed between the plates,

C' = k*C

C' = 25* 1.85*10^(-10) = 46.33*10^(-10) F

New charge,

Q' = 46.33*10^(-10) * 600 = 27.8*10^(-7) C

Additional charge,

Q = 27.8*10^(-7) - 1.112*10^(-7)

Q = 2.67*10^(-6) C

Q = 2.67 uC