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Initially, a single capacitanceC1is wired to a battery. Then capacitanceC2is added in parallel. Are (a) the potential difference acrossC1and (b) the chargeq1onC1now more than, less than, or the same as previously? (c) Is the equivalent capacitanceC12ofC1andC2more than, less than, or equal toC1? (d) Is the charge stored onC1and C2 together more than, less than, or equal to the charge stored previously onC1?

Short Answer

Expert verified

a) The potential difference is the same than previously on C1.

b) The q1on C1now the same as previously.

c) The equivalent capacitance C12of C1and C1more than C1.

d) The charge stored on C1and C1together more than the charge stored previously on C1.

Step by step solution

01

The given data

The capacitance C1and C1are in parallel.

02

Understanding the concept of a parallel capacitors connection

Using the properties of parallel circuits and using Eq.25-19 and 25-1, we can predict the potential difference across capacitor 1, the charge 1 on capacitor 1, and compare it with its previous value. Also, we can find the equivalent capacitance of capacitor 1 and capacitor 2 using the corresponding formula. Thus, this can determine whether it is more than, less than, or equal to capacitor 1, and the charge stored on capacitor 1 and capacitor 2 together can be found using Eq.25-1 and we can determine whether it is more than, less than, or equal to the charge stored previously on capacitor 1.

Formulae:

The charge within the plates of the capacitor,q=CV …(i)

If capacitors are in parallel, the equivalent capacitanceCeqis given by,

Ceq=C …(ii)

03

(a) Calculation of the potential difference on capacitor 1

Considering the properties of the parallel circuits, the potential remains unchanged in the parallel capacitors.

Therefore, we can note that the potential difference is the same than previously across C1.

04

(b) Calculation of charge on capacitor 1

The properties of parallel circuits give the charges that remain the same in the circuit considering equation (i).

Therefore, the charge q1on C1is now the same as previous.

05

(c) Calculation of the equivalent capacitance

From equation (ii), the equivalent capacitance of this parallel connection is given by:

C12=C1+C2

Therefore, capacitance is increased after parallel arrangement. This implies that, the equivalent capacitor C12of C1andC2 is more than C1.

06

(d) Calculation of the charge stored on both the capacitors

From calculations of part (c), we can note that iftheequivalent capacitorC12ofC1andC2 is more thanC1 then from equation (i), we get that the charge is more.

Hence, the charge on C1and C2combined is also more than the previously stored charge on C1.

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Most popular questions from this chapter

You have many 2.0μFcapacitors, each capable of withstanding 200 Vwithout undergoing electrical breakdown (in which they conduct charge instead of storing it). How would you assemble a combination having an equivalent capacitance of (a)0.40μFand (b)1.2μF, each combination capable of withstanding 1000 V?

In Fig. 25-28, find the equivalent capacitance of the combination. Assume that C1is10.0μF,C2is5.00μF,andC3is4.00μF

Figure 25-20 shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated (top plate positive). After the switches have been closed, in which circuit (if any) will the charge on the left-hand capacitor (a) increase, (b) decrease, and (c) remain the same?

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.

A potential difference of 300 V is applied to a series connection of two capacitors of capacitances C1=2.00μFand C2=8.00μF.What are (a) charge q1 and (b) potential difference V1on capacitor 1 and (c) q2and (d) V2 on capacitor 2? The chargedcapacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) q1 , (f) V1 , (g) q2 , and (h) V2? Suppose, instead,the capacitors charged in part (a) are reconnected with plates of oppositesigns wired together. What now are (i) q1, ( j)V1 , (k)q2 , and (l)V2?

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