Option 4 : 200 μF

**Concept:**

Energy stored by the inductor (E_{L}) is given by,

\(E_L=\frac{1}{2}L{I^2}\)

Energy stored by a capacitor (E_{C}) is given by,

\(E_C =\frac{1}{2}C{V^2}\)

**Calculation:**

**Given:** L = 2 H , I = 4 A, V_{C } = 400 V

From the above concept,

\(\frac{1}{2}L{I^2} = \frac{1}{2}C{V^2}\)

\(L{I^2} = C{V^2}\)

\(2 \times {\left( 4 \right)^2} = C{\left( {400} \right)^2}\)

\(C = \frac{{32}}{{160000}}\)

\(C = 200\mu F\)

Option 2 : 8 J

__Concept__:

Energy stored in capacitor:

- A capacitor is a device to store energy.
- The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
- The work done in charging the capacitor is stored as its electrical potential energy.
- The energy stored in the capacitor is

\(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where Q = charge stored on the capacitor,

U = energy stored in the capacitor,

C = capacitance of the capacitor and

V = Electric potential difference

When capacitors are connected in** parallel,** the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in **series,** the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

**Calculations:**

C_{eq} = 4 μF, V = 400

When capacitances are connected in series

then, \(\frac{1}{{{C_{eq}}}} = \frac{5}{C}\)

C = 20 μF

When capacitances are connected in parallel

C_{eq}' = 20 × 5 = 100 μF

\(U = \frac{{100 \times {{10}^{ - 6}} \times 400 \times 400}}{2}\)

=** 8 J**

Option 1 : 0.6

**Concept:**

In a parallel plate capacitor, the energy density is given by

\(\vec E = \frac{1}{2}\frac{{{{\left| D \right|}^2}}}{\varepsilon }\)

D is the flux density

ε is permittivity

**Calculation:**

Flux density \(\vec D = 15\;\mu C/{m^2}\)

= 15 × 10^{-6} C/m^{2}

Energy density \(\vec E = 20\;J/{m^3}\)

To find relative permittivity ε_{r}:

We know that,

\(\Rightarrow 20\frac{J}{{{m^3}}} = \frac{1}{2} \times \frac{{{{\left( {15\; \times \;{{10}^{ - 6}}} \right)}^2}}}{{\frac{1}{{36\pi }}\; \times \;{{10}^{ - 9}}{\varepsilon _r}}}\)

\(\Rightarrow {\varepsilon _r} = \frac{{15\; \times \;15\; \times \;{{10}^{ - 12}}\; \times \;36\pi }}{{2\; \times \;20\; \times\; {{10}^{ - 9}}}}\)

⇒ ε_{r} = 0.635 ≈ 0.6

The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is 𝐸 when the plate separation is 𝑑. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2𝑑.

At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects?

Option 1 : 2𝐸

**Concept: **

Capacitance with the separation is r is given by \({{\rm{C}}} = \frac{{{\rm{\epsilon A}}}}{{{\rm{r}}}}\)

**Application:**

Capacitance when the separation is 2d.

\({{\rm{C}}_1} = \frac{{{\rm{\epsilon A}}}}{{2{\rm{d}}}}\)

Let the charge on plates be Q coulomb. Then, the energy of the capacitor with separation 2d is

\(\begin{array}{l} {{\rm{E}}_1} = \frac{{{{\rm{Q}}^2}}}{{2{{\rm{C}}_1}}}\\ {{\rm{E}}_1} = \frac{{{{\rm{Q}}^2}}}{{2{{\rm{C}}_1}}} = \frac{{{{\rm{Q}}^2}2{\rm{d}}}}{{2{\rm{\epsilon A}}}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{{\rm{\epsilon A}}}} \end{array}\)

The energy of the capacitor with separation d is

Now, \({\rm{C}} = \frac{{{\rm{A}}}}{{\rm{d}}} \Rightarrow {\rm{E}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{2{\rm{\epsilon A}}}}\)

\(\Rightarrow {\rm{E}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{2{\rm{\epsilon A}}}}\)

Thus, we see \({\rm{E}} = \frac{1}{2}{{\rm{E}}_1}\)

\(\Rightarrow {{\rm{E}}_1} = 2{\rm{E}}\)Option 4 : 200 μF

**Concept:**

Energy stored by the inductor (E_{L}) is given by,

\(E_L=\frac{1}{2}L{I^2}\)

Energy stored by a capacitor (E_{C}) is given by,

\(E_C =\frac{1}{2}C{V^2}\)

**Calculation:**

**Given:** L = 2 H , I = 4 A, V_{C } = 400 V

From the above concept,

\(\frac{1}{2}L{I^2} = \frac{1}{2}C{V^2}\)

\(L{I^2} = C{V^2}\)

\(2 \times {\left( 4 \right)^2} = C{\left( {400} \right)^2}\)

\(C = \frac{{32}}{{160000}}\)

\(C = 200\mu F\)

Option 3 : 40 J/m^{3}

**Concept:**

Electrical energy density = total energy per unit volume

\({W_e} = \frac{1}{2}\;\vec Q \cdot \vec E = \frac{1}{2}\varepsilon {E^2}\;J/{m^3}\)

Where ε → Medium permittivity

E → Electric field or potential gradient.

**Calculation:**

Given that the medium is air.

ε = 8.854 × 10^{-12} F/m

Potential gradient, E = 30 × 10^{3} × 10^{2} V/m

= 3 × 10^{6} V/m

\({W_e} = \frac{1}{2} \times 8.854 \times {10^{ - 12}} \times 3 \times {10^6} \times 3 \times {10^6} = 40\;J/{m^3}\)

Option 2 : 8 J

__Concept__:

Energy stored in capacitor:

- A capacitor is a device to store energy.
- The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
- The work done in charging the capacitor is stored as its electrical potential energy.
- The energy stored in the capacitor is

\(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where Q = charge stored on the capacitor,

U = energy stored in the capacitor,

C = capacitance of the capacitor and

V = Electric potential difference

When capacitors are connected in** parallel,** the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in **series,** the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

**Calculations:**

C_{eq} = 4 μF, V = 400

When capacitances are connected in series

then, \(\frac{1}{{{C_{eq}}}} = \frac{5}{C}\)

C = 20 μF

When capacitances are connected in parallel

C_{eq}' = 20 × 5 = 100 μF

\(U = \frac{{100 \times {{10}^{ - 6}} \times 400 \times 400}}{2}\)

=** 8 J**

A spherical balloon of radius 'a' is charged. The energy density in the electric field at point P shown in the figure given below is W. If the balloon is inflated to a radius 'b' without altering its charge, what is the energy density at P?

Option 4 : W

__Concept__:

Gauss Law states that the net electric flux through a closed surface enclosing a volume equals the charge enclosed by it.

\(\mathop \oint \nolimits_s \overset{\rightharpoonup}{E}.\overset{\rightharpoonup}{{ds}} = \frac{Q}{{{\varepsilon _o}}}\)

Since D = ϵ0 E, we can write:

\(\mathop \oint \nolimits_s \overline D .\overset{\rightharpoonup}{{ds}} = Q\)

D̅ is the electric flux density.

__Observations__**:**

1) For a Gaussian sphere around point P, the net charge enclosed by it will not change with a change in radius.

2) Since the net outward electric field through any closed surface is equal to the charge enclosed by it, the electric field outside the spherical balloon will not change with the change in its radius.

3) So, the energy density at point P will be the same as W for the inflated radius b of the balloon.

Option 1 : 0.6

**Concept:**

In a parallel plate capacitor, the energy density is given by

\(\vec E = \frac{1}{2}\frac{{{{\left| D \right|}^2}}}{\varepsilon }\)

D is the flux density

ε is permittivity

**Calculation:**

Flux density \(\vec D = 15\;\mu C/{m^2}\)

= 15 × 10^{-6} C/m^{2}

Energy density \(\vec E = 20\;J/{m^3}\)

To find relative permittivity ε_{r}:

We know that,

\(\Rightarrow 20\frac{J}{{{m^3}}} = \frac{1}{2} \times \frac{{{{\left( {15\; \times \;{{10}^{ - 6}}} \right)}^2}}}{{\frac{1}{{36\pi }}\; \times \;{{10}^{ - 9}}{\varepsilon _r}}}\)

\(\Rightarrow {\varepsilon _r} = \frac{{15\; \times \;15\; \times \;{{10}^{ - 12}}\; \times \;36\pi }}{{2\; \times \;20\; \times\; {{10}^{ - 9}}}}\)

⇒ ε_{r} = 0.635 ≈ 0.6

Option 3 : ML^{-1}T^{-2}

__CONCEPT:__

- Energy Density: It is defined as the amount of energy stored in a given system or a region of space per unit volume.
- Energy density = Energy stored / volume.
- The SI Unit of Energy Density is Joule/metre3 or J/m3. (∵ Energy is measured in joule and volume in m3).
- The SI base unit is - kg·m−1s−2
- Mathematically it is written as

\(Energy\;density = \frac{{Energy}}{{volume}}=\frac{1}{2}{{ε }_{0}}{{E}^{2}}\)

__CALCULATION__:

- Mathematically it is written as

\(Energy\;density = \frac{{Energy}}{{volume}}=\frac{1}{2}{{ε }_{0}}{{E}^{2}}\)

As we know the dimension the energy (E) = [ML^{2}T^{-2}]

The dimension of the volume (V) = [L^{3}]

∴ The dimension of energy density is

\(\Rightarrow\frac{1}{2}{{ε }_{0}}{{E}^{2}}=\frac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}]\)

- Hence option 3 is the answer.