Question

Chapter 18-Homework Practice Problem 18.10 Now let's examine the effects of placing a dieletric film Part C- Practice Problem: between the plates of a capacitor. The plates of an air led parallel-plate capacitor each harve area 2.00 x 10 em and are 1.00 em apart. The capacilor is connected to a power difference of V6 = 3.00 kV The capacitor is then disconnected from the power supply without any charge being lost from its plates. After the capacitor has been disconnected, a sheet of insulating plastic material is inserted between the plates, completely filling the space between them When this is done, the potenbal difference Find the electric field between the plates before the dielectric is inserted Express your answer in volts per meter to three significant figures. 100 V/m between the plates decreases to 1.00 kV, while the charge remains constant (a) What is the capacitance of the capacitor before and after the dielectric is inserted? (b) What is the dielectric constant of the plastic? (c) What is the electric field between the plates bofore and ater the dielectric is inserted? Submit Incorrect Try Again: 3 attempts remaining Figure t012) Part D Practice Problem: Fund the electrk fold between the plates after the diefectnc is inserted Express your answer in volts per meter to three significant figures. d-100 cR

Answer 1

Initially the potential difference between the plates of capacitor is $V_0=3.00\,kV$, when dielectric is inserted potential difference between the plates of capacitor is reduced to $V=1.00\,kV$

$V=\frac{V_0}{k}\Rightarrow k=\frac{V_0}{V}=\frac{3.00\,k}{1.00\,k}=3$

Dielectric constant of material inserted is $k=3$

$d=1.00\,cm=1.00*10^{-2}\,m$

Part C)

Potential difference between the plates of capacitor ${V_0}=7.00\,kV={7.00*10^3}\,V$

Electric field between the plates before the dielectric is inserted is $E=\frac{V_0}{d}=\frac{7.00*10^3}{1.00*10^{-2}}=7*10^5\,V/m$

Part D)

After the dielectric is inserted, the electric field remains the same as it depends on ${V_0},d$ . The values of ${V_0}$ and $d$ do not change.

The polarization of dielectric tries to decrease the electric field but increase in charge on plates by battery maintains the potential difference $V_0$ constant, there by keeping electric field constant.

Hence, electric field between the plates after the dielectric is inserted is$E=\frac{V_0}{d}=\frac{7.00*10^3}{1.00*10^{-2}}=7*10^5\,V/m$