A parallel-plate capacitor with plate area A = 2.0 m² and plate separation d = 3.0 mm is connected to a 35-V battery (Fig. 17–51a). (a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength K = 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17–51b). What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor? A = 2.0 m² }d = 3.0 mm 35 V (a) FIGURE 17-51 35 V K= 3.2 3.0 mm Problem 96. (b)

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A parallel-plate capacitor with plate area A = 2.0 m² and
plate separation d = 3.0 mm is connected to a 35-V battery
(Fig. 17–51a). (a) Determine the charge on the capacitor, the
electric field, the capacitance, and the energy stored in
the capacitor. (b) With the capacitor still connected to the
battery, a slab of plastic with dielectric strength K = 3.2
is placed between the plates of the capacitor, so that the gap is
completely filled with the dielectric (Fig. 17–51b). What are
the new values of charge,
electric field, capacitance,
and the energy stored in
the capacitor?
A = 2.0 m²
}d = 3.0 mm
35 V
(a)
FIGURE 17-51
35 V
K= 3.2 3.0 mm
Problem 96.
(b)
Transcribed Image Text:A parallel-plate capacitor with plate area A = 2.0 m² and plate separation d = 3.0 mm is connected to a 35-V battery (Fig. 17–51a). (a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength K = 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17–51b). What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor? A = 2.0 m² }d = 3.0 mm 35 V (a) FIGURE 17-51 35 V K= 3.2 3.0 mm Problem 96. (b)
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