Question

le stockroom for EAMIPLE 18.7 Properties of a parallel-plate capacitor et's apply a potential difference or voltage across a parallel-plate capacitor and calculate the resulting and electric field. Suppose the plates of a parallel-plate capacitor are 5.00 mm apart and 20 Video Tutor Solution charge and el Ia potential difference of 10.0 kV is applied across the capacitor, compute (a) the capacitance, (bi) dhe and 2.00 m in on each plate, and (e) the magnitude of the electric field in the region between the plates. harge on SOLUTION SET UP Figure 18.20 shows our sketch. SOLVE Part (a): To find the capacitance, we use Equation 18.16, A 2.00 m2 -10.0 kV d-500 mm which expresses capacitance in terms of the area and separation distance of the plates: ▲ FIGURE 18.20 Our sketch for this problem. 10-12 F/m)(2.00 5.00 × 10-3 m foA (8.85 × ) Solving for gives : = 3.54 × 10-9 F-0.00354 μF. Q = Cv,b = (3.54 × 10-9 F)(1.00 × 104 v) = 3.54 × 10-5 C = 35.4pc. Part (b): Now that we know the capacitance, we can use it and the voltage to find the charge on the plates. Remember that capacitance is charge divided by voltage: C O/Vab

Answer 1

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