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Given a 7.4 pF air filled capacitor, you are asked to convert it to a capacitor that can store up to 7.4μJwith a maximum potential difference of 652 v. Which dielectric in table should you use to fill the gap in the capacitor if you do not allow for a margin of error?

Short Answer

Expert verified

The dielectric material that needs to be filled within the gap is Pyrex.

Step by step solution

01

The given data

a) Capacitance value without dielectric, C0=7.4×10-12F

b) Maximum potential difference, V = 652 V

c) Energy stored in capacitor with dielectric, U=7.4×10-6J

02

Understanding the concept of the capacitance with dielectric

If the space between the plates of a capacitor is completely filled with a dielectric material, the capacitance is C increased by a factor k, called the dielectric constant, which is characteristic of the material. In a region that is completely filled by a dielectric, all electrostatic equations containingε0 must be modified by replacingε0 withkε0.

We can use the relation between capacitance without dielectric and with a dielectric material and the equation for energy in terms of capacitance and potential difference to find the dielectric constant. Using the product of the capacitance of the capacitor without dielectric (or with air, ) and the dielectric constant of the material, we can get the value of the capacitance with dielectric. Thus, by substituting this in the energy equation, we can find the value of the dielectric constant.

Formulae:

The capacitance relation with dielectric to without dielectric,Ck=k×C0 …(i)

Here,Ck is capacitance with dielectric, k is dielectric constant,C0 is capacitance without dielectric.

The energy stored between the capacitors,U=12×CkV2 …(ii)

Here, U is the stored energy,Ck is the capacitance with dielectric and V is the potential difference between two plates.

03

Calculation of the dielectric constant

Using the given data in the substituted equation of (i) in equation (ii), we can get the value of the dielectric constant as follows:

C0×k=2UV2k=2UC0V2=2×7.4×10-6J7.4×10-12F×652V2=4.7

Hence, comparing this dielectric value, we get that the material is Pyrex.

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Most popular questions from this chapter

A certain parallel plate capacitor is filled with a dielectric for which k = 5.5. The area of each plate is 0.034m2, and the plates are separated by 2.0 mm.The capacitor will fail (short out and burn up) if the electric field between the plates exceeds200kNC. What is the maximum energy that can be stored in the capacitor?

Figure 25-22 shows an open switch, a battery of potential difference, a current-measuring meter A, and three identical uncharged capacitors of capacitance C. When the switch is closed and the circuit reaches equilibrium, what is (a) the potential difference across each capacitor and (b) the charge on the left plate of each capacitor? (c) During charging, what net charge passes through the meter?

In Fig. 25-31, a 20.0 Vbattery is connected across capacitors of capacitancesC1=C6=3.00μFandC3=C5=2.00C2=2.00C4=4.00μFWhat are (a) the equivalent capacitanceCeqof the capacitors and (b) the charge stored byCeq? What are (c)V1and (d)role="math" localid="1661748621904" q1of capacitor 1, (e)role="math" localid="1661748675055" V2and (f)q2of capacitor 2, and (g)V3and (h)q3of capacitor 3?


Two parallel plates of area 100cm2are given charges of equal magnitudes 8.9x10-7C, but opposite signs. The electric field within the dielectric material filling the space between the plates is1.4x106Vm(a)Calculate the dielectric constant of the material(b)Determine the magnitude of charge induced on each dielectric surface.

Figure 25-18 shows plots of charge versus potential difference for three parallel-plate capacitors that have the plate areas and separations given in the table. Which plot goes with which capacitor?

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