A parallel-plate capacitor has a capacitance ofC o = 5.40 pF when there is air b

Question

A parallel-plate capacitor has a capacitance ofCo = 5.40 pF when there is air betweenthe plates. The separation between the plates is 1.50 mm. a.) What is the maximum magnitude of charge thatcan be placed on each plate if the electric field in the regionbetween the plates is not to exceed3.00×104 V/m? b.) A dielectric with a dielectric constant of3.20 is inserted between the plates of the capacitor, completelyfilling the volume between the plates. Now what is the maximummagnitude of charge on each plate if the electric field between theplates is not to exceed 3.00×104 V/m? A parallel-plate capacitor has a capacitance ofCo = 5.40 pF when there is air betweenthe plates. The separation between the plates is 1.50 mm. a.) What is the maximum magnitude of charge thatcan be placed on each plate if the electric field in the regionbetween the plates is not to exceed3.00×104 V/m? What is the maximum magnitude of charge thatcan be placed on each plate if the electric field in the regionbetween the plates is not to exceed3.00×104 V/m? b.) A dielectric with a dielectric constant of3.20 is inserted between the plates of the capacitor, completelyfilling the volume between the plates. Now what is the maximummagnitude of charge on each plate if the electric field between theplates is not to exceed 3.00×104 V/m? A dielectric with a dielectric constant of3.20 is inserted between the plates of the capacitor, completelyfilling the volume between the plates. Now what is the maximummagnitude of charge on each plate if the electric field between theplates is not to exceed 3.00×104 V/m?

Explanation / Answer

C0=5.4x10-12F d=1.5x10-3m a)V=Ed and V=Q/C hence Q=CEd Since E=3.00x10-4 is not to be exceeded, wehave Q=C0Ed=(5.4x10-12)(3.00x10-4)(1.5x10-3)=2.43x10-18       (maximum magnitude of charge) b)Now C=3.2C0 (with dielectric). Substituting inQ=CEd we have Q=7.776x10-18

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