Q: Suppose a charge +Q1 is given to the positive plate and a charge −Q2 to the negative plate of a capacitor. What is the "charge on the capacitor"?

Given:Charge on the positive plane =+Q1 Charge on the negative plate =−Q2 To calculate: Charge on the capacitorLet ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y , respectively.Let q be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.Let the area of the plates be A and the permittivity of the free space be ϵ0 .Now, to determine q in terms of Q1 and Q2 , we need to apply Gauss's law to calculate the electric field due to all four faces of the capacitor at point P.Also, we know that the electric field inside a capacitor is zero.Electric field due to face I at point P,E1=2∈0AQ1−q Electric field due to face II at point P,E2=2ϵ0A+q Electric field due to face III at point P,E3=2ϵ0A−q Electric field due to face IV at point P,E4=−(2ϵ0A−Q2+q) (Negative sign is used as point P lies on the LHS of face IV .)Since point P lies inside the conductor,E1+E2+E3+E4=0∴Q1−q+q−q−(−Q2+q)=0⇒q=2Q1+Q2 Thus, the charge on the capacitor is 2Q1+Q2 , which is the charge on faces II and III.

Question 1

Suppose a charge +Q1​ is given to the positive plate and a charge −Q2​ to the negative plate of a capacitor. What is the "charge on the capacitor"?

Accepted Answer

Given:Charge on the positive plane =+Q1​ Charge on the negative plate =−Q2​ To calculate: Charge on the capacitorLet ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y , respectively.Let q be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.Let the area of the plates be A and the permittivity of the free space be ϵ0​ .Now, to determine q  in terms of Q1​ and Q2​ , we need to apply Gauss's law to calculate the electric field due to all four faces of the capacitor at point P.Also, we know that the electric field inside a capacitor is zero.Electric field due to face I at point P,E1​=2∈0​AQ1​−q​ Electric field due to face II at point P,E2​=2ϵ0​A+q​ Electric field due to face III at point P,E3​=2ϵ0​A−q​ Electric field due to face IV at point P,E4​=−(2ϵ0​A−Q2​+q​) (Negative sign is used as point P lies on the LHS of face IV .)Since point P lies inside the conductor,​E1​+E2​+E3​+E4​=0∴Q1​−q+q−q−(−Q2​+q)=0⇒q=2Q1​+Q2​​​ Thus, the charge on the capacitor is 2Q1​+Q2​​ , which is the charge on faces II and III.

Question 2

As C=(V1​)Q , can you say that the capacitance C is proportional to the charge Q?

Accepted Answer

No. Since capacitance is a proportionality constant, it depends neither on the charge on the plates nor on the potential. It only depends upon the size and shape of the capacitor and on the dielectric used between the plates.The formula that shows its dependence on the size and shape of the capacitor is as follows:C=dϵσ​A​ Here, A is the area of the plates of the capacitor and d  is the distance between the plates of the capacitor.

Question 3

A hollow metal sphere and a solid metal sphere of equal radii are given equal charges. Which of the two will have higher potential?

Accepted Answer

The potential of a metal sphere is directly proportional to the charge q given to it and inversely proportional to its radius r .i.e. V=4πϵ0​rq​ Since both the spheres are conductors with the same radius and charge, the charge given to them appears on the surface evenly. Thus, the potential on the surface or within the sphere will be the same, no matter the sphere is hollow or solid.

Question 4

The plates of a parallel-plate  capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

Accepted Answer

It is given that the plates of the capacitor have the same charges. In other words, they are at the same potential, so the potential difference between them is zero.Let us consider that the charge on face II is q so that the induced charge on face III  is −q and the distribution is according to the figure. Now, if we consider Gaussian surface ABCD, whose faces lie inside the two plates, and calculate the field at point P due to all four surfaces, it will be​E1​=2ϵ0​AQ−q​E2​=2ϵ0​Aq​E3​=−2ϵ0​Aq​E4​=−2ϵ0​AQ+q​ (It is -ve because point P is on the left side of face IV.) ​ Now, as point P lies inside the conductor, the total field must be zero.

Question 5

A capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charges? If no, what other information is needed?

Accepted Answer

No. This information is not sufficient. Since the charge is proportional to the potential difference across the capacitor, we need to know the potential difference applied across the capacitor.q ∝ V ⇒ q = CVHere, q is the charge, V is the potential difference applied and C is the proportionality constant, i.e. capacitance.

Question 6

The dielectric constant decreases if the temperature is increased. Explain this in terms of polarization of the material.

Accepted Answer

The amount of polarisation can be understood as the extent of perfect alignment of the molecules of a dielectric with an external electric field. The more aligned the molecules are with the external magnetic field, the more is the polarisation and the more will be the dielectric constant.But with increase in temperature, the thermal agitation of the molecules or the randomness in their alignment with the field increases.Thus, we can say that increase in temperature results in reduced polarisation and reduced dielectric constant.

Question 7

When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?

Accepted Answer

As the energy of the system decreases, the change in the energy is negative.Force is defined as a negative rate of change of energy with respect to distance.F=−∂x∂U​ So, as the energy decreases, the force due to the electric field of the capacitor increases when the dielectric is dragged into the capacitor.

Question 8

A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is:

Accepted Answer

ZeroSince the net charge enclosed by the Gaussian surface is zero, the total flux of the electric field through the closed Gaussian surface enclosing the capacitor is zero.HereΦ=∬​E.ds=ϵ0​q​=0 Φ = Electric flux q = Total Charge enclosed by the Gaussian surface .

Question 9

Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be:

Accepted Answer

Since the voltage gets added up when the capacitors are connected in series, the voltage of the combination is 2 V . Also, the capacitance of a series combination is given byCmet ​1​=C1​1​+C2​1​ ​Cnet ​= Net capacitance of the combination C1​=C2​=C∴Cnet ​=2C​​

Question 10

If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be:

Accepted Answer

2C and VIn a parallel combination of capacitors, the potential difference across the capacitors remain the same, as the right-hand-side plates and the left-hand-side plates of both the capacitors are connected to the same terminals of the battery. Therefore, the potential remains the same, that is, V.For the parallel combination of capacitors, the capacitance is given by​Ceq ​=C1​+C2​ Here, C1​=C2​=C∴Ceq​=2C​

Question 11

The equivalent capacitance of the combination shown in the figure is _________ .

Accepted Answer

Since the potential at point A is equal to the potential at point B, no current will flow along arm AB. Hence, the capacitor on arm AB will not contribute to the circuit. Also, because the remaining two capacitors are connected in parallel, the net capacitance of the circuit is given byCeq​=C+C=2C

Question 12

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will:

Accepted Answer

Remain unchanged.The force between the plates is given by F=2ϵ0​Aq2​  Since the capacitor is isolated, the charge on the plates remains constant. We know that the charge is conserved in an isolated system. Thus, the force acting between the plates remains unchanged.

Question 13

The energy density in the electric field created by a point charge falls off with the distance from the point charge as:

Accepted Answer

r41​ Energy density U is given byU=21​∈0​E2 The electric field created by a point charge at a distance r is given by E=4πϵ0​r2q​ On putting the above form of E in eq. 1 , we getU=21​∈0​(4πϵ0​r2q​)2 Thus, U is directly proportional to r41​ .

Question 14

A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q, and Q be the charges appearing on the positive and negative plates respectively.

Accepted Answer

Q+​=Q−​ The charge induced on the plates of a capacitor is independent of the area of the plates.∴Q+​=Q−​

Question 15

A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates . The capacitance now becomes _________ .

Accepted Answer

(d) Indeterminate.The thin metal plate inserted between the plates of a parallel-plate capacitor of capacitance C connects the two plates of the capacitor; hence, the distance d between the plates of the capacitor reduces to zero. It can be observed that the charges on the plates begin to overlap each other via the metallic plate and hence begin to conduct continuously.Mathematically,C=d∈0​A​ In this case, d=0 .∴C=∞

Question 16

The following figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.

Accepted Answer

(c) C1​<C​2​.Region AB shows the potential difference across capacitor C​1​  and region CD shows the potential difference across capacitor C2​ . Now, we can see from the graph that region AB is greater than region CD. Therefore, the potential difference across capacitor C1​  is greater than that across capacitor C2​. ∵ Capacitance, C=VQ​ ∴ C1​       (Q remainsthesameinseriesconnection.)

Question 17

Two metal plates having charges Q, −Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will:

Accepted Answer

IncreaseOil between the plates of the capacitor acts as a dielectric. We know that the electric field decreases by a factor of 1/K of the original field when we insert a dielectric between the plates of a capacitor (K is the dielectric constant of the dielectric). So, if the oil is pumped out, the electric field between the plates will increase, as the dielectric has been removed.

Question 18

Two metal spheres of capacitance C1​ and C2​ carry some charges. They are put in contact and then separated. The final charges Q1​  and Q2​  on them will satisfy.

Accepted Answer

Q2​Q1​​=C2​C1​​ When the spheres are connected, charges flow between them until they both acquire the same common potential V . The final charges on the spheres are given by'Q1​=C1​V and Q2​=C2​V Q2​Q1​​=C2​VC1​V​=C2​C1​​

Question 19

Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are:

Accepted Answer

2 µF, 18 µFThe minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:​C1​=61​+61​+61​=21​⇒C=2μF​ The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows: C=6+6+6=18μF

Question 20

The capacitance of a capacitor does not depend on:

Accepted Answer

The charges on the platesThe capacitance of a capacitor is given by C=dϵ0​A​ Here, A is the area of the plates of the capacitor and d is the distance between the plates.So, we can clearly see that the capacitance of a capacitor does depend on the size and shape of the plates and the separation between the plates; it does not depend on the charges on the plates.

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