Informative line

Dielectric

Practice work done when dielectric slab is inserted between the plates of capacitor and effect on electric field when dielectric is inserted between capacitor. Learn breakdown voltage for parallel combination of capacitors.

Effect of Dielectric on Capacitance

Definition : Dielectric is a non - conducting material such as rubber, glass or waxed paper.

Case 1 :

•  Consider a parallel plate capacitor having capacitance $$C_0$$ , connected  to a battery to charge upto $$Q_0$$ and then disconnected it from battery.
• The potential difference across the capacitor is given as

$$\Delta V_0= \dfrac{Q_0}{ C_0}$$

or,  $$C_0 = \dfrac{Q_0}{ \Delta V_0}$$

Case 2 :

A dielectric slab is inserted between the capacitor plates.

• Due to dielectric, the potential difference across the capacitor is decreased by a factor K.

$$\Delta V = \dfrac{\Delta V_0}{K}$$

Hence, the capacitance will be

$$C = \dfrac{Q_0}{\Delta V}$$

$$\Rightarrow C = \dfrac{Q_0}{\Delta V_0/K}$$

$$\Rightarrow C= \dfrac{KQ_0}{\Delta V_0}$$

$$C=KC_0$$   where $$C_0 = \dfrac{Q_0}{\Delta V_0}$$

Note:  $$Q_0$$ remains the same even after inserting the slab as slab does not allow charge to move through it.

• Since, after inserting the slab (dielectric) , potential is reduced .

$$\Delta V < \Delta V_0$$

Hence,  K > 1

•  The factor K is known as dielectric constant of the material.

A capacitor having capacitance $$C_0=6\,\mu F$$ is charged through a battery $$\mathcal{E}= 120\,V$$ . Calculate the capacitance after a dielectric slab having dielectric constant $$K= 2$$ , is inserted between the plates.

A $$6\,\mu F$$

B $$12\,\mu F$$

C $$18\,\mu F$$

D $$4\,\mu F$$

×

New capacitance $$(C)$$ when dielectric slab is inserted between capacitor plate is given as

$$C=C_0K$$    [where , $$K$$ = constant , $$C_0$$ = capacitance without dielectric]

Given:  $$C_0=6\,\mu F$$ , $$K=2$$

$$C=6\times10^{-6}\times2$$

$$C=12\,\mu F$$

A capacitor having capacitance $$C_0=6\,\mu F$$ is charged through a battery $$\mathcal{E}= 120\,V$$ . Calculate the capacitance after a dielectric slab having dielectric constant $$K= 2$$ , is inserted between the plates.

A

$$6\,\mu F$$

.

B

$$12\,\mu F$$

C

$$18\,\mu F$$

D

$$4\,\mu F$$

Option B is Correct

Dielectric Strength and Breakdown Voltage

Polarization of Dielectric

• When a dielectric slab is placed in an external electric field, the dielectric gets polarized.
• Due to polarization , positive charge is accumulated at left end and negative charge is accumulated at right end of the dielectric.
• But as a whole, dielectric is still neutral .
• The dielectric creates an electric field $$\vec E_{induced}$$ inside it which is in opposite direction of electric field $$(\vec E_0)$$.

Effect on Electric Field when Dielectric is Inserted between Capacitor (Disconnected from Battery)

• In the absence of dielectric the direction of electric field is as shown in figure.

• When dielectric is inserted between plates of parallel plate capacitor, an electric field is induced inside the dielectric due to polarization which is opposite to the electric field $$(E_0)$$ applied.

• Hence, the net electric field is in direction , as shown in figure.

$$\vec E_{net} = E_0- E_{induced}$$

Note : The direction  of net electric field is still from positive plate to negative plate but its strength is weaker than $$E_0$$.

• Due to insertion of dielectric , the electric field weakens by a factor (K).

$$\vec E_{net} =\dfrac {\vec E_0} {K}$$

Dielectric Strength

It is the maximum value of electric field at which the dielectric gets fully ionized and behaves as conductor.

Breakdown Voltage

• The Maximum value that can be applied to a capacitor without causing a discharge .
• The voltage at which the capacitor starts to discharge after the maximum value, is known as breakdown voltage.

Relation between Breakdown Voltage and Dielectric Strength

$$V_{Breakdown} = E_{ds} × d$$

where Eds =  Dielectric Strength

d = Separation between plates

Calculate the charge at the breakdown voltage (Vbd) of a capacitor having capacitance C with the separation between the plates of area A is 'd'.

A $$\dfrac{AV_{bd}}{K\epsilon_0}$$

B $$\dfrac{AV_{bd}K}{\epsilon_o}$$

C $$AV_{bd} \,K \epsilon_o$$

D $$\dfrac{AV_{bd}\,\epsilon_o}{d}$$

×

The value of electric field at breakdown voltage is known as dielectric strength of dielectric.

Relation between breakdown voltage and dielectric strength is given as

$$V_{bd} = E_{ds} \cdot d$$

$$E_{ds} = \dfrac{V_{bd}}{d}$$

$$\dfrac {\sigma}{\epsilon_o}$$ = $$\dfrac{V_{bd}}{d}$$               $$\left [\therefore \,E_{ds} = \dfrac{\sigma}{\epsilon_o} \right]$$

$$\dfrac {Q}{A\,\epsilon_o}$$ = $$\dfrac{V_{bd}}{d}$$                  $$\left [\because \,\sigma = \dfrac{Q}{A} \right]$$

$$Q = \dfrac{A V_{bd}\,\epsilon_o}{d}$$

Calculate the charge at the breakdown voltage (Vbd) of a capacitor having capacitance C with the separation between the plates of area A is 'd'.

A

$$\dfrac{AV_{bd}}{K\epsilon_0}$$

.

B

$$\dfrac{AV_{bd}K}{\epsilon_o}$$

C

$$AV_{bd} \,K \epsilon_o$$

D

$$\dfrac{AV_{bd}\,\epsilon_o}{d}$$

Option D is Correct

Change in Properties of Capacitor Due to Insertion of Dielectric Slab

• Consider a parallel plate capacitor of capacitance C, connected to a battery of e.m.f. $$\mathcal{E}$$.
• A dielectric slab of dielectric constant K is inserted between plates of capacitor.
• Charge before the insertion of dielectric slab,

$$Q_1 = C × \mathcal{E}$$

• Charge after the insertion of dielectric slab,

$$Q'_1 = K C \, \mathcal{E}$$

• Charge supplied through battery ,

$$\Delta Q = Q_f - Q_i$$

$$\Delta Q = KC \mathcal{E} - C \mathcal{E}$$

$$\Delta Q = C \mathcal{E} \, (K- 1)$$

• Capacitance of parallel  plate capacitor before insertion of slab,

$$C_1=C$$

• Capacitance of parallel  plate capacitor after insertion of slab,

$$C' = KC$$

• Work done by battery before insertion of slab,

$$W_1 = Q × \mathcal{E}$$

$$W_1 = C\mathcal{E} × \mathcal{E}$$

$$W_1 = C\mathcal{E} ^2$$

• Work done by battery after insertion of slab,

$$W'_1 = \Delta Q × \mathcal{E}$$

$$W'_1 = (K-1)\,C\mathcal{E} × \mathcal{E}$$

$$W'_1 = (K-1)\,C\mathcal{E} ^2$$

• Electric field in capacitor remains constant  when a dielectric is inserted when capacitor connected to battery.

$$E = \dfrac{V}{d}$$

V and d are constant.

$$\Rightarrow$$ E is constant.

• Energy stored in capacitor before insertion of slab,

$$U_1 = \dfrac{1}{2} CV^2$$

• Energy stored in capacitor after insertion of slab,

$$U'_1 = \dfrac{1}{2} K CV^2$$

• Change in energy $$\Delta U = U'_1 - U_1 = \dfrac{1}{2} CV^2(K-1)$$        ,

Table : When capacitor connected to battery,

 S.No. Capacitor Before Insertion of Slab Capacitor After Insertion of Slab 1. Capacitance $$\to$$ C KC 2. Charge $$\to$$ Q KQ 3. Work done by battery $$\to$$ W (K-1)W 4. Energy stored $$\to$$ U KU 5. Electric field $$\to$$ E E

A parallel plate capacitor of capacitance $$C= 50\,\mu F$$  is connected to a battery of $$\mathcal{E} = 100V$$. A dielectric slab of dielectric constant K = 3 is inserted between plates of capacitor. Find the extra charge flown through the battery.

A $$5\,mC$$

B $$20\,mC$$

C $$15\,mC$$

D $$10\,mC$$

×

Charge before insertion of dielectric slab

$$Q = C \mathcal{E}$$

$$Q = 50 × 10^{-6} × 100$$

$$Q = 5$$ $$mC$$

Charge after insertion of dielectric slab

$$Q' = KCV$$

$$Q' = 3 × 50 × 10 ^{–6} × 100$$

$$Q' = 15 \,mC$$

Extra charge $$\Delta Q$$ flow through battery

$$\Delta Q = Q_f - Q_i = Q' -Q$$

$$\Delta Q = 15\, mC - 5\,mC$$

$$\Delta Q = 10\,mC$$

A parallel plate capacitor of capacitance $$C= 50\,\mu F$$  is connected to a battery of $$\mathcal{E} = 100V$$. A dielectric slab of dielectric constant K = 3 is inserted between plates of capacitor. Find the extra charge flown through the battery.

A

$$5\,mC$$

.

B

$$20\,mC$$

C

$$15\,mC$$

D

$$10\,mC$$

Option D is Correct

Disconnection

• Consider a capacitor of capacitance C, charged with battery.
• The battery is disconnected when capacitor is fully charged.
• Now, a dielectric slab of dielectric constant K, is inserted into it.
• The capacitance before insertion of dielectric  = C
• The capacitance after insertion of dielectric  = KC

• Charge will remain constant after disconnection of battery since there is no circuit to move the charge .

Hence, there is no impact of dielectric insertion on charge.

• Energy stored in capacitor before insertion of dielectric

$$U_i = \dfrac{Q^2}{2C}$$

• Energy stored in capacitor after insertion of dielectric

$$U_f = \dfrac{Q^2}{2KC}$$

• Change in energy stored in capacitor due to dielectric insertion

$$\Delta U = U_f - U_i$$

$$\Delta U = \dfrac {Q^2}{2KC} - \dfrac{Q^2}{2C}$$

$$\Delta U = \dfrac {Q^2}{2C} \left(\dfrac{1}{K} -1\right)$$

$$\Delta U = \dfrac {Q^2}{2C} \left(\dfrac{1-K}{K} \right)$$

Effect on Electric Field when Dielectric is Inserted between Capacitor (Disconnected from Battery)

• In the absence of dielectric the direction of electric field is as shown in figure.

• When dielectric is inserted between plates of parallel plate capacitor, an electric field is induced inside the dielectric due to polarization which is opposite to the electric field $$(E_0)$$ applied.

• Hence, the net electric field is in direction , as shown in figure.

$$\vec E_{net} = E_0- E_{induced}$$

Note : The direction  of net electric field is still from positive plate to negative plate but its strength is weaker than $$E_0$$.

• Due to insertion of dielectric, the electric field weakens by a factor (K).

$$\vec E_{net} =\dfrac {\vec E_0} {K}$$

Table :  When capacitor disconnected from battery

 Capacitor Before Insertion of Dielectric Capacitor After Insertion of Dielectric Capacitance $$\to$$ C KC Charge   $$\to$$ Q Q (Constant) Work done by battery $$\to$$ W Zero Energy stored $$\to$$ U $$\dfrac{U}{K}$$ Electric field $$\to$$ E $$\dfrac{E}{K}$$

A capacitor of capacitance $$C = 5 \mu F$$ is charged to $$Q = 10 \mu C$$ and then disconnected. Calculate the change in energy stored in capacitor when a dielectric slab of dielectric constant $$K = 2$$ is inserted between the capacitor plate.

A $$5 \,\mu J$$

B $$-5 \,\mu J$$

C $$10 \,\mu J$$

D $$15 \,\mu J$$

×

Energy stored in capacitor before insertion of dielectric

$$U_i = \dfrac{Q^2}{2C}$$

$$U_i = \dfrac{(10 × 10^{-6})^2} {2×5×10^{-6}}$$

$$U_i = 10 \,\mu J$$

Energy stored in capacitor after insertion of dielectric

$$U_f = \dfrac{Q^2}{2KC}$$

$$U_f = \dfrac{(10 × 10^{-6})^2} {2×2×5×10^{-6}}$$

$$U_f = 5 \, \mu J$$

Change in energy stored in capacitor due to dielectric insertion

$$\Delta U = U_f - U_i$$

$$\Delta U = 5 \mu J - 10 \mu J$$

$$\Delta U = -5\mu J$$

A capacitor of capacitance $$C = 5 \mu F$$ is charged to $$Q = 10 \mu C$$ and then disconnected. Calculate the change in energy stored in capacitor when a dielectric slab of dielectric constant $$K = 2$$ is inserted between the capacitor plate.

A

$$5 \,\mu J$$

.

B

$$-5 \,\mu J$$

C

$$10 \,\mu J$$

D

$$15 \,\mu J$$

Option B is Correct

Breakdown Voltage for Series Combination

• Consider  two capacitors , having capacitance C1 and C2 respectively, are connected in series, as shown in figure.
• The breakdown voltage of capacitor Cis V1  and breakdown voltage of capacitor Cis V2.

Steps to Calculate Breakdown Voltage of Combination

Step 1:

Find the equivalent capacitance of combination.

$$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

or, $$\dfrac{1}{C_{eq}} = \dfrac{C_1 +C_2}{C_1C_2}$$

or, $$C_{eq} = \dfrac{C_1 C_2}{C_1 +C_2}$$

Step 2:

Calculate the maximum charge on each capacitor.

$$(Q_1)_{max} = C_1 ×V_1$$

$$(Q_2)_{max} = C_2 ×V_2$$

Step 3:

• For series combination, charge remains same.
• So, compare (Q1)max and (Q2)max  .
• The charge with less value is limiting charge and if more than this charge flows in the combination, it will destroy one of the capacitor.

Step 4:

Breakdown voltage of this combination is

$$V_{br} = \dfrac {Q_{limit}}{C_{eq}}$$

Two capacitors of capacitance $$C_1 = 2 \,\mu F$$ and $$C_2 = 1 \,\mu F$$ having breakdown voltage $$V_{br1} = 3\,kV$$ and $$V_{br2} = 4\,kV$$ respectively,  are connected in series ,as shown in figure. Calculate the breakdown voltage of this combination.

A 6 kV

B 4 kV

C 3 kV

D 5 kV

×

Equivalent capacitance (Ceq) of the combination

$$\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

$$\Rightarrow \dfrac{1}{C_{eq}} = \dfrac{1}{2×10^{-6}} + \dfrac{1}{1×10^{-6}}$$

$$C_{eq} = \dfrac{2}{3} \mu F$$

Maximum charge on capacitor C1

$$(Q_1)_{max} = V_{br1} × C_1$$

$$\Rightarrow (Q_1)_{max} = 3×10^3 × 2×10^{-6}$$

$$(Q_1)_{max} = 6\,mC$$

Maximum charge on capacitor C2

$$(Q_2)_{max} = V_{br2} × C_2$$

$$\Rightarrow (Q_2)_{max} = 4×10^3 × 1×10^{-6}$$

$$(Q_2)_{max} = 4\,mC$$

For series combination , charge is same.

Since, $$(Q_1)_{max} >(Q_2)_{max}$$

Hence, (Q2) max   will be the limiting charge because more than this flow of charge will destroy C.

$$\therefore Q_{limit} = 4 \,mC$$

Breakdown voltage $$(V_{br})_{C_1 C_2}$$ for combination

$$(V_{br})_{C_1 C_2} = \dfrac{Q_{limit}}{C_{eq}}$$

$$(V_{br})_{C_1 C_2} = \dfrac{4×10^{-3}} {\dfrac{2}{3}× 10^{-6}}$$

$$(V_{br}) _ {C_1C_2} = 6\,kV$$

Two capacitors of capacitance $$C_1 = 2 \,\mu F$$ and $$C_2 = 1 \,\mu F$$ having breakdown voltage $$V_{br1} = 3\,kV$$ and $$V_{br2} = 4\,kV$$ respectively,  are connected in series ,as shown in figure. Calculate the breakdown voltage of this combination.

A

6 kV

.

B

4 kV

C

3 kV

D

5 kV

Option A is Correct

Breakdown Voltage for Parallel Combination of Capacitors

• Consider two capacitors of capacitance C1 and C2 , having breakdown voltage V1 and V respectively such that   V1 < V2 ,  are  connected in  parallel.
• Since in parallel , voltage remains same . Hence, breakdown voltage of the combination will be the lowest breakdown voltage of any of the capacitor connected in parallel.
• So, breakdown voltage of the combination is V1 volt.

Two capacitors of capacitance $$C_1 = 2 \,\mu F$$   and $$C_2 = 3 \,\mu F$$ , have breakdown voltage $$V_1 = 3 \,kV$$ and  $$V_2 = 5 \,kV$$ respectively, are connected in parallel. Determine the value of breakdown voltage for this combination.

A 3 kV

B 7 kV

C 5 kV

D 10 kV

×

Since in parallel, voltage remains same . Hence, breakdown voltage of the combination will be the lowest breakdown voltage of any of the capacitor connected in parallel.

Breakdown voltage (V1) of C1 = 3 kV

Breakdown voltage (V2) of C2 = 5 kV

So, V1 < V2

Hence, breakdown voltage of combination

Vbr = 3 kV

Two capacitors of capacitance $$C_1 = 2 \,\mu F$$   and $$C_2 = 3 \,\mu F$$ , have breakdown voltage $$V_1 = 3 \,kV$$ and  $$V_2 = 5 \,kV$$ respectively, are connected in parallel. Determine the value of breakdown voltage for this combination.

A

3 kV

.

B

7 kV

C

5 kV

D

10 kV

Option A is Correct

Work Done when Dielectric Slab is Inserted between the Plates of Capacitor

• Consider a parallel plate capacitor of capacitance C, connected to a battery of e.m.f. $$\mathcal{E}$$.
• A dielectric slab of dielectric constant K is inserted between plates of capacitor.
• Charge before the insertion of dielectric slab

$$Q_1 = C × \mathcal{E}$$

• Charge after the insertion of dielectric slab

$$Q'_1 = K C \, \mathcal{E}$$

• Charge supplied through battery

$$\Delta Q = Q_f - Q_i$$

$$\Delta Q = KC \mathcal{E} - C \mathcal{E}$$

$$\Delta Q = C \mathcal{E} \, (K- 1)$$

• Capacitance of parallel  plate capacitor before insertion of slab

$$C_1=C$$

• Capacitance of parallel  plate capacitor after insertion of slab

$$C' = KC$$

• Work done by battery before insertion of slab.

$$W_1 = Q × \mathcal{E}$$

$$W_1 = C\mathcal{E} × \mathcal{E}$$

$$W_1 = C\mathcal{E} ^2$$

• Work done by battery after insertion of slab

$$W'_1 = \Delta Q × \mathcal{E}$$

$$W'_1 = (K-1)\,C\mathcal{E} × \mathcal{E}$$

$$W'_1 = (K-1)\,C\mathcal{E} ^2$$

• Electric field in capacitor remains constant when a dielectric is inserted and when capacitor connected to battery.

$$E = \dfrac{V}{d}$$

V,d and E are constant.

• Energy stored in capacitor before insertion of slab

$$U_1 = \dfrac{1}{2} CV^2$$

• Energy stored in capacitor after insertion of slab

$$U'_1 = \dfrac{1}{2} K CV^2$$

• Change in energy $$\Delta U = U'_1 - U_1 = \dfrac{1}{2} CV^2(K-1)$$

When capacitor connected to battery

 S.No. Capacitor Before Insertion of Slab Capacitor After Insertion of Slab 1. Capacitance $$\to$$ C KC 2. Charge $$\to$$ Q KQ 3. Work done by battery $$\to$$ W (K-1)W 4. Energy stored $$\to$$ U KU 5. Electric field $$\to$$ E E

• While inserting  the dielectric slab between the capacitor plates connected to battery, the electric field attracts the dielectric slab with force F.
• An external force of equal magnitude F should be applied so that the plate moves slowly (no acceleration).

By work - Energy theorem

$$W_b + W_{ext} = \Delta U + heat$$      ( $$\therefore$$ KE = 0, moving slowly)

• Since the dielectric slab is inserted slowly so, heat will be zero.

$$(K- 1) C \mathcal{E}^2 + W _{ext} = \dfrac{1}{2} (K-1) C \mathcal{E}^2$$

$$W_{ext} = \dfrac{1}{2} (K-1) C \mathcal{E}^2 - (K-1) C\mathcal{E}^2$$

$$W_{ext} = -\dfrac{1}{2} (K-1) C \mathcal{E}^2$$

A parallel plate capacitor of capacitance $$C=5\,\mu F$$ is connected to a battery of e.m.f. $$\mathcal{E}=100V$$ . A dielectric slab of dielectric constant  $$K =2$$ is now inserted into the gap between plates. Calculate  Work done by battery Work done by external source Change in energy.

A $$40\, mJ, 80\, mJ, 25\, mJ$$

B $$50\, mJ, –25\, mJ, 25\, mJ$$

C $$25\, mJ, 50\, mJ, –25\, mJ$$

D $$60 \,mJ, 30 \,mJ, –20\, mJ$$

×

Charge before insertion of dielectric

$$Q_i = C \,\mathcal{E}$$

$$Q_i = 5× 10^{-6} × 100$$

$$Q_i = 0\cdot5 \,mC$$

Charge after insertion of dielectric

$$Q_f = KC \,\mathcal{E}$$

$$Q_f = 2×5× 10^{-6} × 100$$

$$Q_f = 1\,mC$$

Extra charge supplied by battery

$$\Delta Q = Q_f - Q_i$$

$$\Delta Q = 1\,mC - \cdot\,5\,mC$$

$$\Delta Q = \cdot\,5\,mC$$

Work done by battery

$$W_b = \Delta Q × \mathcal{E}$$

$$W_b = 0\cdot 5\,mC × 100$$

$$W_b = 50\,mJ$$

Energy stored in capacitor before insertion of dielectric

$$U_i = \dfrac{1}{2} C \mathcal{E}^2$$

$$U_i = \dfrac{1}{2} ×5×10^{-6} × (100)^2$$

$$U_i = 25\,mJ$$

Energy stored in capacitor after insertion of dielectric

$$U_f = \dfrac{1}{2} KC \mathcal{E}^2$$

$$U_f = \dfrac{1}{2} × 2×5× 10^{-6} × (100)^2$$

$$U_f =50 \,mJ$$

Change in energy stored in capacitor due to insertion of dielectric slab

$$\Delta U = U_f - U_i$$

$$\Delta U = 50\,mJ - 25 \, mJ$$

$$\Delta U = 25 \, mJ$$

By  Work - Energy Theorem

$$W_b + W_{ext} = \Delta U$$

Where , $$W_b$$  = Work done by battery

$$W_{ext }$$= External work done

$$\Delta U=$$ Change in kinetic energy

So, $$50\, mJ + W_{ext} = 25 \,mJ$$

$$W_{ext} =\, – 25\, mJ$$

A parallel plate capacitor of capacitance $$C=5\,\mu F$$ is connected to a battery of e.m.f. $$\mathcal{E}=100V$$ . A dielectric slab of dielectric constant  $$K =2$$ is now inserted into the gap between plates. Calculate  Work done by battery Work done by external source Change in energy.

A

$$40\, mJ, 80\, mJ, 25\, mJ$$

.

B

$$50\, mJ, –25\, mJ, 25\, mJ$$

C

$$25\, mJ, 50\, mJ, –25\, mJ$$

D

$$60 \,mJ, 30 \,mJ, –20\, mJ$$

Option B is Correct

The diagram shows four capacitor with their capacitance and breakdown voltage, as mentioned. What  should be the maximum value of the external e.m.f. source such that no capacitor breaks down ?

A $$2\,kV$$

B $$1.\,5\,kV$$

C $$3\,kV$$

D $$6\,kV$$

×

For Branch 1

Equivalent capacitance $${C_{eq_1}}$$ of the combination

$$\dfrac{1}{C_{eq_1}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

$$\Rightarrow\dfrac{1}{C_{eq_1}} = \dfrac{1}{2×10^{-6}} + \dfrac{1}{1×10^{-6}}$$

$$C_{eq_1} = \dfrac{2}{3} \mu F$$

Maximum charge on capacitor C1

$$(Q_1)_{max} = V_{br1} × C_1$$

$$\Rightarrow(Q_1)_{max} = 3×10^3 × 2×10^{-6}$$

$$(Q_1)_{max} = 6\,mC$$

Maximum charge on capacitor C2

$$(Q_2)_{max} = V_{br2} × C_2$$

$$\Rightarrow(Q_2)_{max} = 4×10^3 × 1×10^{-6}$$

$$(Q_2)_{max} = 4\,mC$$

For series combination , charge is same.

Since, $$(Q_1)_{max} >(Q_2)_{max}$$

Hence, (Q2max   will be the limiting charge because more than this flow of charge will destroy C.

$$\therefore Q_{limit} = 4 \,mC$$

Breakdown voltage $$(V_{br})_{C_1 C_2}$$ for combination

$$(V_{br})_{C_1 C_2} = \dfrac{Q_{limit}}{C_{eq_1}}$$

$$(V_{br})_{C_1 C_2} = \dfrac{4×10^{-3}} {\dfrac{2}{3}× 10^{-6}}$$

$$(V_{br}) _ {C_1C_2} = 6\,kV$$

For branch 2

Equivalent capacitance (Ceq 2) of combination

$$\dfrac{1}{(C_{eq})_2} = \dfrac{1}{3×10^{-6}} +\dfrac{1}{6×10^{-6}}$$

$$(C_{eq})_2 = 2 \, \mu F$$

Maximum charge on capacitor C3

$$(Q_3)_{max} = V_{br3} × C_3$$

$$(Q_3)_{max} = 1 × 10^3 × 3×10^{-6}$$

$$(Q_3)_{max} = 3\,mC$$

Maximum charge on capacitor C4

$$(Q_4)_{max} = V_{br4} × C_4$$

$$(Q_4)_{max} = 2 × 10^3 × 6×10^{-6}$$

$$(Q_4)_{max} = 12\,mC$$

For series combination , charge is same .

Since, $$(Q_4)_{max} > (Q_3)_{max}$$

Hence, (Q3)max  will be the limiting charge because more than this flow of charge will destroy capacitor C3

$$\therefore$$ $$Q_{limit} = 3\,mC$$

Breakdown voltage $$(V_{br})_ {C3 C4}$$ for combination

$$(V_{br})_ {C3 C4} = \dfrac{Q_{limit}}{(C_{eq})_2}$$

$$(V_{br})_ {C3 C4} = \dfrac{3\,mC}{2\,\mu F}$$

$$(V_{br})_ {C3 C4} = 1.5\,kV$$

For parallel combination of branch 1 and branch 2

Since in parallel , voltage remains same. Hence, breakdown voltage of the combination will be the lowest breakdown voltage of any of the capacitor connected in parallel.

Breakdown voltage of branch 1

$$(V_{br})_{C1C2} = 6\,kV$$

Breakdown voltage of branch 2

$$(V_{br})_{C3C4} = 1\cdot5\,kV$$

So, $$(V_{br})_{C1C2} > (V_{br})_{C3C4}$$

Hence, breakdown voltage of combination

$$(V_{br})_{eq} = 1\cdot5\,kV$$

So, the maximum value of e.m.f. source such that no capacitor breaks, is

$$\mathcal{E} = 1\cdot5 \,kV$$

The diagram shows four capacitor with their capacitance and breakdown voltage, as mentioned. What  should be the maximum value of the external e.m.f. source such that no capacitor breaks down ?

A

$$2\,kV$$

.

B

$$1.\,5\,kV$$

C

$$3\,kV$$

D

$$6\,kV$$

Option B is Correct