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tester [92]
3 years ago
10

A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the vol

tage across its plates if the space between them becomes filled with mica? (the dielectric constant for mica is 5.4 and for air is 1.0006.)
Physics
1 answer:
andrezito [222]3 years ago
7 0

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling C_0 the capacitance of the capacitor in air, the charge Q, the capacitance C_0 and the voltage (V_0=391 V) are related by

C_0 =\frac{Q}{V_0} (1)


when the source is disconnected the charge Q remains on the capacitor.


When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

C=k C_0 = 5.4 C_0

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

V=\frac{Q}{C}=\frac{Q}{5.4 C_0}

And since Q=C_0 V_0, substituting into the previous equation, we find:

V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V



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