Physics M.C.Q [1M]

3. Electric potential

Explanation:

Electric potential may be defined as the amount of work done in moving a unit positive charge from infinity to a given point.

$\text{V}=\frac{\text{W}}{\text{q}}$

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. $\frac{\text{C}}{2}\ \text{and}\ \text{2V}$

Explanation:

Since the voltage gets added up when the capacitors are connected in series, the voltage of the combination is 2V.

Also, the capacitance of a series combination is given by

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

Here,

C_​net​ = Net capacitance of the combination

$\text{C}_1=\text{C}_2=\text{C}$

$\therefore\text{C}_{\text{net}}=\frac{\text{C}}{2}$

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 $\frac{1}{\text{r}^4}.$

3. $8\mu\text{F}$

Explanation:

In the given figure,

Equivalent capcitance between A and B.

C_{eg ​}= C₁ ​+ C_2​

$=4μ\text{F}+4μ\text{F}$

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

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

4. $\text{Zero}.$

Explanation:

Since 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.

$\phi=\oint\text{E.ds}=\frac{\text{q}}{\in_0}=0$

Here,

$\phi=$ Electric flux

q = Total charge enclosed by the Gaussian surface.

2. If V = 0, E must be zero.

Explantion:

Electric field is negative gradient of electric potential.

$\text{E}=-\text{grad}(\text{V})$

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

If $\text{E}=0$

$-\frac{\text{dV}}{\text{dr}}=0$

This implies

V = constant.

A constant can be zero or a quantifiable number.

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

3. $\text{r}$

1. 1.8F

Explanation:

Since we know that capacitance$=\frac{\text{charge}}{\text{voltage}}$

Therefore capacitance$=\frac{9}{5}=1.8\text{f.}$

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

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

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.