(Solved): Problem 1: Dielectric materials in capacitors A capacitor is constructed with an air gap of 2mm b ...

Problem 1: Dielectric materials in capacitors A capacitor is constructed with an air gap of $2\mathrm{mm}$ berween its parallel plates: A. What is the capacitance per square meter for such a capacitor? Use this result to calculate the capacitance if the plates are squares of side $0.6\mathrm{m}$. B. If the capacitor stores a charge $Q$, find expressions for the voltage difference between the plates and for the magnitude $E_0$ of the electric field between the plates in terms of $Q$ only. C. If you were to halve the separation between the plates, determine for each quantity listed if it increases or decreases (and by which factor), or if it remains unchanged: 1. the capacitance of the capacitor, 2. the voltage difference between the plates, 3. the magnitude $E_0$ of the electric field between the plates. The space between the plates is entirely filled with paper, which has a dielectric constant of $?=2.9$. D. What is the capacitance per square meter for such a capacitor? Use this result to calculate the capacitance if the plates are squares of side $0.6\mathrm{m}$. E. If the capacitor stores a charge $Q$, find expressions for the voltage difference between the plates and for the magnitude $E_x$ of the electric field between the plates in terms of $Q$ only. 2. Use your previous result to calculate the capacitance of the capacitor. (Hint: recall that $V=Q/C$ ) H. The capacitor is equivalent to two capacitors connected as pictured, where the mid-equipotential line has been replaced by conductors (also equipotential regions). Use your results from $\mathrm{Cl}$ and $\mathrm{F}1$ to determine the capacitance of this combination.