Physics M.C.Q (1 Marks)

4. $\frac{\text{F}}{8}$

Explanation:

The electric field due to dipole at distance r is $\text{E}\propto\frac{1}{\text{r}^3}.$

Thus force on charge is $\text{F}=\text{qE}$

$\Rightarrow\text{F}\propto\frac{1}{(2\text{r})^3}=\frac{\text{F}}{8}.$

4. 60

Explanation:

The equivalent capacitance of the two $6μ\text{F} $ and $6μ\text{F} $ capacitors in series is $3μ\text{F} $.

Hence the potential across the two capacitors, original $3μ\text{F} $ capacitor and the equivalent $3μ\text{F} $ capacitor is divided equally.

4. 27

Explanation:

Place three 200V, $6μ\text{F},$ capacitors in series to get 1 equivalent 600V, $2μ\text{F},$ capacitor. Now place 9 of these equivalent 600V, $2μ\text{F},$capacitors in parallel to obtain an equivalence of 18μF at 600 Volts. All this requires a total of 27 $6μ\text{F},$ capacitors. Nine rows connected in parallel with 3 capacitors connected in series in each row.

2. For any x for a given z.
3. For any y for a given z.
4. On the x-y plane for a given z.

We know, the electric field intensity E and electric potential V are

$\text{E}=-\frac{\text{dV}}{\text{dr}}$

Electric potential decreases inf the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.

The electric field in z-direction suggest that equipotential surfaces are in x-y plane. Therefore the potential is a constant for any x for a given z, for any y for a given z and on the x-y plane for a given z.

1. 4

Explanation:

We can obtain an equivalence capacitance of $5μ\text{F},$ by connecting minimum  4 capacitance of each $2μ\text{F},$only in the way as shown in the figure.

3. Area of the plate is increased.

Explanation:

Hint:- Check the dependence of capacitance on certain quantities.

In a parallel plate capacitor, the capacitance is $\text{C}=\frac{\text{k}\in_0\text{A}}{\text{d}};$

Where, k is the dielectric constant, $\in_0$​ is the permittivity constant, A is the area of the conduction and d is the distance between plates.

From here we can see C is directly proportional to k, $\in_0$​,A and inversely proportional to d.

3. The work done is the same in Fig. (i) Fig. (ii) and Fig. (iii).

Key concept: For a given charge distribution, locus of all points or regions for which the electric potential has a constant value are called equipotential regions. Such equipotential can be surfaces, volumes or lines. Regarding equipotential surface the following points should be kept in mind:

1. The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.
2. The direction of electric field is perpendicular to the equipotential surfaces or lines.
3. The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.
4. For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field lines.
5. A metallic surface ofany shape is an equipotential surface.
6. Equipotential surfaces can never cross each other.
7. The work done in moving a charge along an equipotential surface is always zero.

As the direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential than other equipotential surface maintained at low electrostatic potential. Hence direction of electric field is from B to A in all three cases.
The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction opposite to electric field. Thus, the work done by the electric field on the charge will be negative. We know

$\text{W}_\text{electrical}=-\Delta\text{U}=-\text{q}\Delta\text{V}=\text{q}(\text{V}_\text{Intial}-\text{V}_\text{final})$

Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.

3. 2C and V

Explanation:

In 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

C_eq = C₁ + C₂

Here,

C₁ = C₂ = C

$\therefore$ C_eq = 2C

2. Q₊ = Q_-

Explanation:

The charge induced on the plates of a capacitor is independent of the area of the plates.

$\therefore$ Q₊ = Q_-

1. Both 1 and 2 are true, 2 is not correct explanation of 1

Explanation:

Consider a capacitor with charge density $σ.$

The potential between its two plates is given by $\frac{\sigma\text{d}}{\in_0}$

When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.

However, this is nothing to the dielectric strength of the dielectric.

3. C₁ < C₂

Explanation:

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. 

$\because$ Capacitance, $\text{C}=\frac{\text{Q}}{\text{V}}$

$\therefore\text{C}_1<\text{C}_2$ (Q remains the same in series connection).

3. 1 Joule/ Coulomb

Explanation:

The volt is a measure of electric potential. One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, s, and A)

$1\text{volte}=\frac{1\text{joule}}{\text{Coulomb}}$

2. The charge on the capacitor.

Explanation:

When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor.

Thus, the net effect is a reduced electric field.

Also, as the potential is proportional to the field, the potential decreases and so does the stored energy U, which is given by

$\text{U}=\frac{\text{qV}}{2}$

Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.

3. Remain unchanged.

Explanation:

The force between the plates is given by

$\text{F}=\frac{\text{q}^2}{2\in_0\text{A}}$

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.

2. Decreases

Explanation:

When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field.

The reduction in the electric field results in a decrease in potential difference.

$\text{V}=\text{Ed}=\frac{\text{E}_0\text{d}}{\text{k}}=\frac{\text{V}_0}{\text{k}}$

1. Statement is correct

Explanation:

The given statement is correct. Volt or Joules per Coulomb are SI unit of Potential (V) i.e, work done per unit charge to move a charge from a reference point to a certain point.

And electric field E is negative of the gradient of electric potential (V) i.e,$\text{E}=\frac{\text{-dv}}{\text{dr}}$

2. The electric field is zero.
3. There can be no charge inside the region.

We know, the electric field intensity E and electric potential V are dV related as $\text{E}=-\frac{\text{dV}}{\text{dr}}$

or we can write $|\text{E}|=-\frac{\Delta\text{V}}{\Delta\text{r}}$

The electric field intensity E and electric potential V are related as E = 0 and for V = constant, $\frac{\text{dV}}{\text{dr}}=0$ this imply that electric field intensity E = 0.

If some charge is present inside the region then electric field cannot be zero at that region, for this V = constant is not valid.

1. Continuously increases.

Explanation:

$\text{V}=\frac{\text{KQ}}{\text{r}}$

V → Electric Potential

$\text{V}_\text{p}=\frac{\text{KQ}}{\text{x}}+\frac{\text{KQ}}{\text{r}-\text{x}}=\frac{\text{KQ}\text{r}}{\text{x}(\text{r}-\text{x})}$

$\frac{\text{dvp}}{\text{dx}}=\frac{-\text{kQr}(\text{r}-2\text{x})}{\big(\text{r}(\text{r}-\text{x})\big)^2}=0$

$\text{r}=2\text{x},\ \text{x}=\frac{\text{r}}{2}$

$\text{V}_\text{Pmin}=\frac{\text{KQ}}{\frac{\text{r}}{2}}+\frac{\text{KQ}}{\frac{\text{r}}{2}}=\frac{4\text{KQ}}{\text{r}}$ at $\Big(\text{x}=\frac{\text{r}}{2}\Big)$

1. Charge Q

Explanation:

When the capacitor is kept at a voltage, it gains charge.

Now when the system is isolated, the charge present on capacitor cannot change because of law of conservation of charge.

$\therefore$ Charge always remains constant in isolated systems.

4. Equal and opposite charges will appear on the two faces of the metal plate.

Explanation:

The capacitance of the capacitor in which a dielectric slab of dielectric constant K, area A and thickness t is inserted between the plates of the capacitor of area A and separated by a distance d is given by:

$\text{C}=\frac{\in_0\text{A}}{(\text{d}-\text{t})+\big(\frac{\text{t}}{\text{K}}\big)}$

Since it is given that the thickness of the sheet is negligible, the above formula reduces to $\text{C}=\frac{\in_0\text{A}}{\text{d}}.$ In other words, there will not be any change in the electric field, potential or charge.

Only, equal and opposite charges will appear on the two faces of the metal plate because of induction due to the presence of the charges on the plates of the capacitor.

4. 25000

Explanation:

The electric field in between the plates in the presence of dielectric is

$\text{E}=\frac{\text{V}}{\text{Kd}}=\frac{100}{4\times10^{-3}}=25000\text{V/m}$

3. In A : V remains same and hence Q changes.
4. In B : Q remains same and hence V changes.

In case A, key K is kept closed and plates of capacitors are moved apart using insulating handle (i.e. distance between plates is increasing). As capacitance $\text{C}=\Big[\frac{(\text{K}\in_0\text{A})}{\text{d}}\Big]=\text{C}\propto\text{d}$ (separation between plates) so, capacitance will decrease & amount of charge stored will also decrease (as Q = CV). Here, there will be no change in potential difference.

In case B, key K is opened and plates of capacitors are moved apart using insulating handle, by conservation of charge, charge stored by the capacitor remains same. Plates of capacitance are moving apart so capacitance will decrease and with the decreases of capacitance, potential difference V increases $\Big(\text{as V}=\frac{\text{Q}}{\text{C}}\Big)$.

1. Increase.

Explanation:

Oil between the plates of the capacitor acts as a dielectric. We know that the electric field decreases by a factor of $\frac{1}{\text{K}}$ 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.

1. Increases.

Explanation:

Electric Potential Energy $=\text{q}\text{Dv}$

$=\text{q}(\text{v}_\text{f}-\text{v}_\text{i})$

If positive charge is shifted from a Low potential region to a High-Potential region, then electric Potential Energy increases.

1. A

Explanation:

$\text{E}=\text{E}_0\hat{\text{i}}$ (Given)

(Potential at A):-

$\text{V}_\text{A}=\big(\text{E}_0\hat{\text{i}}\times\text{a}\hat{\text{i}}\big)=\text{E}_09$

(Potential atB):-

$\text{V}_\text{B}=\big(\text{E}_0\hat{\text{i}}\times\text{a}\hat{\text{i}}\big)=0$

(Potential at C):-

$\text{V}_\text{C}=\big(\text{E}_0\hat{\text{i}}\times(-\text{a})\text{i}\big)=\text{E}_0\text{a}$

(Potential atD):-

$\text{V}_\text{D}=(\text{E}_0\hat{\text{i}}\times(-\text{a})\hat{\text{j}})=0$

4. Increases the capacity of the condenser.

Explanation:

If a dielectric medium of dielectric constant K is filled completely between the plates then capacitance increases by K times

$\text{C}=\frac{\text{K}\in_0\text{A}}{\text{d}}$

$⇒\text{C}′=\text{KC.}$

4. Q' must be smaller than Q.

Explanation:

The relation between the induced charge Q' and the charge on the capacitor Q is given by:

$\text{Q}'=\text{Q}\Big(1-\frac{1}{\text{K}}\Big)$

Here, K is the dielectric constant that is always greater than or equal to 1.

So, we can see that for K > 1, Q' will always be less than Q.

2. 40 eV

Explanation:

W_AB ​= W_AC ​+ W_CB​

WCB​ should be zero, because in moving from C to B, we always move perpendicular to field. Hence, force applied by field and displacement will be at 90^∘.

So work done in BC will be 0

W_AC ​= −e(V_{C​ -} V_A​)

V_C​ - V_A ​= -E × AC = -10 × 4 = -40

$\therefore$ W_AB​ = 40eJ = 40eV

3. Remain almost the same.

Explanation:

Dynamic resistance (r_P) is given by,

$\text{r}_\text{P}=\frac{\Delta\text{V}_\text{P}}{\Delta\text{i}_\text{P}}$

The triode is operating in the linear region. Therefore,

Change in the value of voltage = Change in the value of current

So, (r_P) will remain almost the same.

2. Decrease in the potential difference across the plates and reduction in the stored energy but no change in the charge on the plates.

Explanation:

If a dielectric slab of dielectric constant K is filled in between the plates of a capacitor after charging the capacitor (i.e., after removing the connection of battery with the plates of capacitor) the potential difference between the plates reduces to $\frac{1}{\text{K}}$​ times and the potential energy of capacitor reduces to$\frac{1}{\text{K}}$times but there is no change in the charge on the plates.

1. Decreases

Explanation:

In general capacitance of parallel plate capacitor is given by:

$\text{C}=\frac{\text{K}\in_0\text{A}}{\text{d}}$

Where C is capacitance, k is relative permittivity of dielectric material, ϵ_0​ is permittivity of free space constant, A is area of plates and d is distance between them. Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material.  Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.

3. $\frac{\text{k}_1\text{k}_2(\text{d}_1+\text{d}_2)}{(\text{k}_1\text{d}_1+\text{k}_2\text{d}_2)}$.

Here the system can be considered as two capacitors C₁ and C₂ connected in series as shown in figure.

The capacitance of parallel plate capacitor filled with dielectric block has thickness d₁ and dielectric constant K₂ is given by

$\frac{1}{\text{C}_\text{eq}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}\Rightarrow\ \text{C}_\text{eq}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}$

$\text{C}_\text{eq}=\frac{\frac{\text{K}_1\in_0\text{A}}{\text{d}_1}}{\frac{\text{K}_1\in_0\text{A}}{\text{d}_1}}\frac{\frac{\text{K}_2\in_0\text{A}}{\text{d}_2}}{\frac{\text{K}_2\in_0\text{A}}{\text{d}_2}}=\frac{\text{K}_1\text{K}_2\in_0\text{A}}{\text{K}_1\text{d}_2+\text{K}_2\text{d}_1}\ ...(\text{i})$

We can write the equivalent capacitance as

$\text{C}=\frac{\text{K}\in_0\text{A}}{\text{d}_1+\text{d}_2}\ ...(\text{ii})$

On comparing (i) and (ii) we have

$\text{K}=\frac{\text{k}_1\text{k}_2(\text{d}_1+\text{d}_2)}{(\text{k}_1\text{d}_2+\text{k}_2\text{d}_1)}$.

2. $\frac{1}{\text{r}^2}$

4. $\frac{1}{\text{r}^4}$

Explanation:

Energy density U is given by

$\text{U}=\frac{1}{2}\in_0\text{E}^2\ \dots(1)$

The electric field created by a point charge at a distance r is given by

$\text{E}=\frac{\text{q}}{4\pi\in_0\text{r}^2}$

On putting the above form of E in eq. 1, we get

$\text{U}=\frac{1}{2}\in_0\Big(\frac{\text{q}}{4\pi\in_0\text{r}^2}\Big)^2$

Thus, U is directly proportional to $\frac{1}{\text{r}^4}.$

4. Indeterminate.

Explanation:

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,

$\text{C}=\frac{\in_0\text{A}}{\text{d}}$

In this case, d = 0.

$\therefore\text{C}=\infty$

3. $\text{r}$

4. $4\sqrt{35}\text{ N}$

2. $\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{C}_1}{\text{C}_2}$

Explanation:

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 = C_​1V and Q₂ = C₂V

$\therefore\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{C}_1\text{V}}{\text{C}_2\text{V}}=\frac{\text{C}_1}{\text{C}_2}$

3. W_A = W_B = W_C

Explanation:

Electric Potential at A 'due to q' þ $\text{V}_\text{A}=\frac{\text{Kq}}{\text{r}}$

Electric Potetial at B 'due to q' þ $\text{V}_\text{B}=\frac{\text{Kq}}{\text{r}}$

& Electric potential at c 'due to q' þ $\text{V}_\text{C}=\frac{\text{Kq}}{\text{r}}$

Work done $=-\text{D}_\text{u}=-\text{q}\text{DV}$ {Let at 'P', V_p = 0}

Here $\text{V}_\text{A}=\text{V}_\text{B}=\text{V}_\text{C}$

The work done is taking a point charge from P to A, B & C is same.

So, $\text{W}_\text{A}=\text{W}_\text{B}=\text{W}_\text{C}$