Figure 25-18 shows a parallel-plate capacitor of plate area A and plate separati

Question

Figure 25-18 shows a parallel-plate capacitor of plate areaA and plate separation d. A potential differenceV0 is applied between the plates. While thebattery remains connected, a dielectric slab of thicknessb and dielectric constant is placed between theplates as shown. Assume A = 103 cm2, d= 1.06 cm, V0 = 60.3 V, b = 0.849 cm,and = 3.06. Calculate (a) the capacitance,(b) the charge (in C) on the capacitor plates,(c) the electric field in the gap, and (d) theelectric field in the slab, after the slab is in place.

Explanation / Answer

for the part A

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