Chapter 4: Q39E (page 811)

A constant potential difference of 12 V is maintained between the terminals of a 0.25-µF, parallel-plate, air capacitor. (a) A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates. When this is done, how much additional charge flows onto the positive plate of the capacitor (see Table 24.1)? (b) What is the total induced charge on either face of the Mylar sheet? (c) What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile this with the increase in charge on the plates, which acts to increasethe electric field.

Short Answer

Expert verified

The additional charge flowing onto the positive plate of the capacitor is $6.3 \times 10^{- 6} C$ .
The total induced charge on either face of the Mylar sheet is $6.3 \times 10^{- 6} C$ .
There is no effect on the electric field between the plates.

Step by step solution

(a) Determination of the additional charge flowing onto the positive plate of the capacitor.

From the mentioned table, the dielectric constant for Mylar is,

$K = 3.1$

The expression for capacitance,

$C = \frac{Q}{V}$

The additional charge,

$∆ Q = n e w c h a r g e - o r i g i n a l c h a r g e = Q - Q_{0} = (K - 1) Q_{0} = (K - 1) C_{0} V_{0} = (2.1) (2.5 \times 10^{- 7} F) (12 V) = 6.3 \times 10^{- 6} C$

Thus, the additional charge is $6.3 \times 10^{- 6} C$ .

(b) Determination of the total induced charge on either face of the Mylar sheet.

The total induced charge is given as,

$Q_{i} = Q (1 - \frac{1}{k}) = (9.3 \times 10^{- 6}) (1 - \frac{1}{3.1}) = 6.3 \times 10^{- 6} C$

Thus, the total induced charge is $6.3 \times 10^{- 6} C$ .

(c) Determination of the effect on the electric field between the plates.

The addition of Mylar doesn’t effect the electric field because from part (a) and (b) it can inferred that the induced charge is equal to the additional charge which will cancel each other. So the original electric field will be maintained.