Physics M.C.Q [1M]

3. S₁ is true, S₂ is also true and S₁ is the cause of S₂.

We know, the electric field intensity E and electric potential V are related $\text{E}=\frac{\text{dV}}{\text{dr}}$.
If electric field intensity  E= 0, then $\frac{\text{dV}}{\text{dr}}=0$.  It means, E = 0 inside the charged conducting sphere causes uniform potential inside the sphere. Hence uniform electrostatic potential 100 V will be at any point inside the sphere.

Important points:

1. The electric field zero does not necessary imply that electric potential is zero. E.g., the electric field intensity at any point inside the charged spherical shell is zero but there may exist non-zero electric potential.
2. If two charged particles of same magnitude but opposite sign are placed, the electric potential at the midpoint will be zero but electric field is not equal to zero.

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

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.

2. The potential difference increases.

Explanation:

Capacitance of a parallel plate capacitor.

$\text{C}=ε0\text{AdC}=ε0\text{Ad}$

A parallel plate capacitor is charged (battery is disconnected) then the plates are pulled apart, the capacitance decreases while the charge remains the same.

$\because$ Potentialdifference = ChargeCapacitance∵Potentialdifference = ChargeCapacitance.

$\therefore$ potential difference increases.

4. May increase or decrease.

Explanation:

When the separation between two charges is increased, the electric potential Energy of charge may incease or decrease.

If Both charge are like charge then electric potential energy of charge decreases.

$\text{U}=\frac{\text{k}\text{q}_1\text{q}_2}{\text{r}}$

If Both charge are unlike charge then electric potential energy of charge increases.

$\text{U}=\frac{-\text{kq}_1\text{q}_2}{\text{r}}$

4. 8μ C.

The capacitor offers infinite resistance in a DC circuit. Here, no current will flow through the capacitor and 10Ω resistance after the capacitor is full charged. 

The potential difference across 10Ω resistance will be zero. It will act like a plain wire.

Current flowing through 2Ω resistance is given by

$\text{I}=\frac{\text{V}}{(\text{R}+\text{r})}=\frac{2.5}{(2+0.5)}=1\text{A}$

Potential difference across 2Ω resistance V = IR = 1 × 2 = 2V

Here, capacitor is connected in parallel with 2 Ω resistance, so it will also have 2V potential difference across it.

The charge on capacitor, q = CV = (2 mF) × 2V = 8 mC.

3. 8 μF

Explanation:

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

When dielectric is added:

$\text{C'}=\frac{\text{K}\in_0\text{A}}{\text{d}}=4\times2\mu\text{F}=8\mu\text{F}$

1. Zero.

Explanation:

In a circle Electric field due tot 'Q' is always perpendicular to the displacement oifthe charge 'q'.

Workdone $=\vec{\text{F}}.\vec{\text{d}}$

$=\big(\text{q}\vec{\text{E}}\big).\vec{\text{d}}=\text{q}\big(\vec{\text{E}}.\vec{\text{d}}\big)=0$

$\therefore\vec{\text{E}}\ \bot\ \vec{\text{d}}$

Workdoen by the Electric field on the rotating charge in one complete revolution is zero.

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.

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,$\text{E}=\frac{\text{-dv}}{\text{dr}}$

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 $\Big(\text{x}=\frac{\text{r}}{2}\Big)$

4. $2\mu\text{F},\ 18\mu\text{F}$

Explanation:

The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:

$\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$

$\Rightarrow\text{C}=2\mu\text{F}$

The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:

$\text{C}=6+6+6=18\mu\text{F}$

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.

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.

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

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.

2. Must be defined as $-\int\limits_\text{A}^\text{B}\text{E}.\text{dl}$.
3. Is zero.

2. Potential difference across the capacitor.
3. Energy of the capacitor.

Explanation:

Because the charge always remains conserved in an isolated system, it will remain the same.

Now,

$\text{V}=\frac{\text{Qd}}{\in_0\text{A}}$

Here, Q, A and d are the charge, area and distance between the plates, respectively.

Thus, as d increases, V increases.

Energy is given by:

$\text{E}=\frac{\text{qV}}{2}$ 

So, it will also increase.

Energy density u, that is, energy stored per unit volume in the electric field is given by:

$\text{u}=\frac{1}{2}\in_0\text{E}^2$

So, u will remain constant with increase in distance between the plates.

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.

2. The charge appearing on the capacitor is greater after the action XWY than after the action XYW.
3. The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
4. The electric field in the capacitor after the action XW is the same as that after WX.

Explanation:

Justification of option (b):

If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of K on inserting a dielectric of a dielectric constant K between the plates of the capacitor. 

Mathematically,

q = Kq₀ 

Here, q₀ and q are the charges without dielectric and with dielectric, respectively.

The amount of charge stored does not depend upon the polarity of the plates.

Thus, the charge appearing on the capacitor is greater after the action XWY than after the action XYZ.

Justification of option (c):

Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is, q = q_​0, because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy U as $\frac{1}{2}\text{q}\in.$ Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor. So, the energy stored in the capacitor will also not change after the action XYW.

However, during the action WXY, the amount of charge that will get stored in the capacitor will get increased by a factor of K, as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of K.

Thus, the electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

Justification of option (d):

The electric field between the plates E depends on the potential across the capacitor and the distanced between the plates of the capacitor.

Mathematically, 

$\text{E}=\frac{\in}{\text{d}}$

In either case, that is, during actions XW and WX, the potential remains the same, that is, $\in.$ Thus, the electric field E remains the same.

Denial of option (a):

During the action XYZ, the battery has to do extra work equivalent to $\frac{1}{2}\text{CV}^2$ to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{CV}^2.$

This extra work done will be dissipated as heat energy. Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is, $\frac{1}{2}\text{CV}^2.$