The capacitance of a parallel-plate capacitor is 24 $\mu$F when the plates are separated by a material of dielectric constant 2.0.If this material is removed,leaving air between the plates,and the separation between the plates is tripled,the capacitance is
A)unchanged
B)16 $\mu$F
C)36 $\mu$F
D)0.14 mF
E)4.0 $\mu$F

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by the formula:
C = εA/d
where ε is the dielectric constant, A is the area of the plates, and d is the distance between them.

When the plates are separated by a material of dielectric constant 2.0, the capacitance is given by:
C1 = 2ε0A/d
where ε0 is the permittivity of free space.

If the material is removed and the separation between the plates is tripled, the capacitance becomes:
C2 = ε0A/3d

Dividing these two equations, we get:
C2/C1 = (ε0A/3d)/(2ε0A/d) = 1/6

Therefore, the capacitance is reduced to 1/6 of its original value, which is 4 F. But since the answer choices are in units of microfarads, we convert 4 F to 4 μF, which is option B.

The Capacitance of a Parallel-Plate Capacitor Is 24 μ\muμ F When the Plates Are Separated by a Material of When

The Capacitance of a Parallel-Plate Capacitor Is 24 $\mu$ F When the Plates Are Separated by a Material of When