Embibe Experts Solutions for Exercise 7: EXERCISE 2.7

A parallel plate air capacitor has capacitance $C.$ Half of space between the plates is filled with di-electric of di-electric constant $K$ as shown in figure. The new capacitance is $C\text{'}.$ Then

A capacitor consists of two metal plates each $10 \mathrm{cm}$ by $20 \mathrm{cm} ;$ they are separated by a $2.0 \mathrm{mm}$ thick insulator with dielectric constant $4.1$ and dielectric strength $6.0\times {10}^7 \mathrm{V} {\mathrm{m}}^{-1}.$ What is the capacitance in $\mathrm{pF}\left({10}^{-12} \mathrm{F}\right)?$

A capacitor consists of two metal plates each $10 \mathrm{cm}$ by $20 \mathrm{cm} ;$ they are separated by a $2.0 \mathrm{mm}$ thick insulator with dielectric constant $4.1$ and dielectric strength $6.0\times {10}^7 \mathrm{V} {\mathrm{m}}^{-1}.$ What minimum voltage will cause a spark in the capacitor.

A parallel plate air capacitor has capacitance $C.$ Now half of the space is filled with a di-electric material with di-electric constant $K$ as shown in figure. The new capacitance is $C.$ Then

A capacitor of capacity $C$ is charged to a potential difference $V$ and another capacitor of capacity $2 \mathrm{C}$ is charged to a potential difference $2 \mathrm{V}.$ The charging batteries are disconnected and the two capacitors are connected to each other in parallel with reverse polarity. The energy lost in this process will be

Two identical capacitors $1$ and $2$ are connected in series. The capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0$ volts. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charge stored in the capacitors before removing the slab and $Q{\text{'}}_1,$ and $Q{\text{'}}_2$ be the values after removing the slab. Then

Two capacitors of capacitances $1.0$ microfarad and $2.0$ microfarad are each charged by being connected across a $5.0$-volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the $2.0$-microfarad capacitor?

A capacitor is charged with a di-electric slab in the space between the plates of capacitor. The battery is disconnected and the di-electric slab is pulled out of the space. Then the force of capacitor upon slab