Answer to Question #106249 in Electricity and Magnetism for havefun7741

Question

[img]https://upload.cc/i1/2020/03/23/RIpb6y.jpg[/img]



R is 4

Answer 1

The capacitance of a flat capacitor is calculated by the formula $C=\epsilon_0\epsilon \cdot \frac{S}{d}$ , where$\epsilon_0=8.854×10^{−12} F⋅m^{−1}$ the electric constant. In the beginning, there was no dielectric in the gap of the capacitor so $\epsilon=1.$ To determine the initial capacitance we should all quantities express in the system SI. $S=20cm^2=20\cdot (10^{-2}m)^2=2.0\cdot 10^{-3}m^2$ ; $d=5mm=5\cdot 10^{-3} m$ $C=8.85\cdot10^{-12}Fm^{-1}\cdot\frac{2.0\cdot 10^{-3}m^2}{5\cdot 10^{-3} m}=3.54\cdot 10^{-12}F$

When the capacitor is charged to a voltage $U=50V$ , a charge will appear on its plates will be

$Q=c\cdot U=3.54\cdot 10^{-12}F\cdot 50V=1.77\cdot10^{-10} Cl$

The electric field between plates find easily

$E=\frac{U}{d}=\frac{50 V}{5\cdot 10^{-3} m}=10^4Vm^{-1}$

When a dielectric is introduced into the capacitor as shown in the figure, it can be considered as two different flat capacitors with the area of the plates doubled in half, connected in parallel.

The capacitanse first (empty) is $C_1=\frac{C}{2}$.The capacitanse second (with dielectric $\epsilon=k=14$ ) is $C_2=14\cdot \frac{C}{2}=7C$. With a parallel connection, the capacitances of the capacitors add up; therefore, the full capacitance of the capacitor after making the dielectric will be $C_{new}=\frac{C}{2}+7C=\frac{15}{2}C=2.65\cdot10^{-11}F$

The electric field inside and outside the slab are the same due to the boundary conditions for the tangential field $E_1{_t}=E_2{_t}$. In our case the side surface of the dielectric is parallel to the direction of the electric field, therefore, its value inside the dielectric coincides with its value in free space.

Electric field we can determine as $E_{new}=\frac{U_{new}}{d}$. As the voltage source is turned off, the charge on the capacitor plates is preserved, so $U_{new}= \frac{Q}{C_{new}}=\frac{1.77\cdot10^{-10} Cl}{2.65\cdot10^{-11}F}=6.7 V$ then

$E_{new}=\frac{U_{new}}{d}=\frac{6.7V}{5\cdot 10^{-3}m}=1.34\cdot 10^3 Vm^{-1}$

Answers:

a)The initial charge the plates is $1.77\cdot10^{-10} Cl$ , initial electric field is $10^4Vm^{-1}$

b)After insertion the slab the capacitance will be $2.65\cdot10^{-11}F$

c)The electric field inside the slab is $1.34\cdot 10^3 Vm^{-1}$

d)The electric field inside and outside the slab are the same, are equal in magnitude, due to the tangential boundary conditions on its side.