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.

**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 \(6\,\mu F\)

B \(12\,\mu F\)

C \(18\,\mu F\)

D \(4\,\mu F\)

- 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 \((\vec E_0)\).

- 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).

\(\vec E_{net} =\dfrac {\vec E_0} {K}\)

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

- 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 E_{ds }= Dielectric Strength

d = Separation between plates

A \(\dfrac{AV_{bd}}{K\epsilon_0}\)

B \(\dfrac{AV_{bd}K}{\epsilon_o}\)

C \(AV_{bd} \,K \epsilon_o \)

D \(\dfrac{AV_{bd}\,\epsilon_o}{d}\)

- Consider two capacitors , having capacitance C
_{1 }and C_{2 }respectively, are connected in series, as shown in figure. - The breakdown voltage of capacitor C
_{1 }is V_{1 }and breakdown voltage of capacitor C_{2 }is V_{2}._{ }

**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 (Q
_{1})_{max }and_{ }(Q_{2})_{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

\(V_{br} = \dfrac {Q_{limit}}{C_{eq}}\)

- Consider two capacitors of capacitance C
_{1}and C_{2}, having breakdown voltage V_{1}and V_{2 }respectively such that V_{1}< V_{2 , }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 V
_{1}volt.

- 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 |

- 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

\(\Delta U = U_f - U_i\)

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

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

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

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

- 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 \(5 \,\mu J\)

B \(-5 \,\mu J\)

C \(10 \,\mu J\)

D \(15 \,\mu J\)

- 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 \(40\, mJ, 80\, mJ, 25\, mJ\)

B \(50\, mJ, –25\, mJ, 25\, mJ\)

C \(25\, mJ, 50\, mJ, –25\, mJ\)

D \(60 \,mJ, 30 \,mJ, –20\, mJ\)

A \(2\,kV\)

B \(1.\,5\,kV\)

C \(3\,kV\)

D \(6\,kV\)