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(Solved): The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential di ...
The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential difference Vo is applied between the plates. While the battery remains connected, a dielectric slab of thickness band dielectric constant K is placed between the plates as shown. Assume A = 107 cm², d - 1.73 cm, Vo = 86.2 V. b=0.826 cm, and K = 1.53. Calculate (a) the capacitance, (b) the charge on the capacitor plates, (c) the electric field in the gap, and (d) the electric field in the slab, after the slab is in place. (a) Number i Units (b) Number i Units Units (c) Number Units (d) Number
The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential difference Vo is applied between the plates. While the battery remains connected, a dielectric slab of thickness b and dielectric constant K is placed between the plates as shown. Assume A-107 cm², d=1.73 cm, Vo 86.2 V, b=0.826 cm, and x = 1.53. Calculate (a) the capacitance, (b) the charge on the capacitor plates, (c) the electric field in the gap, and (d) the electric field in the slab, after the slab is in place. +9 Gaussian surface I 1++ x+ +*+ + Gaussian- surface II (a) Number i Units (b) Number i Units i Units (c) Number Units (d) Number -4- +q²
Part A. We know that capacitance of a capacitor with dielectric slab is given by: C = k1*e0*A0/d0 Now in given case: for air-filled capacitor: C1 = e0*A0/d0, where A0 = A = 107 cm^2 = 0.0107 m^2 d0 = d - b = 1.73 cm - 0.826 cm = 0.9