The capacitance of a parallel plate capacitor:

$\overline{){\mathbf{C}}{\mathbf{=}}\frac{{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}{\mathbf{=}}\frac{{\mathbf{k\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}}$

The charge stored on a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

Energy stored by a capacitor:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{V}}}^{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Q}}{\mathbf{V}}{\mathbf{=}}\frac{{\mathbf{Q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{C}}}$

The electric field in a parallel plate capacitor:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{Q}}{{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{=}}\frac{\mathbf{V}}{\mathbf{d}}}$

**(i)**

C_{0} = ε_{0}A/l

C = kε_{0}A/l

C/C_{0} = kε_{0}A/l ÷ ε_{0}A/l = k

Given a parallel-plate capacitor with plates of area A separated by a distance l. Consider the quantities, (i) capacitance C, (ii) magnitude E of the electric field between the plates, (iii) magnitude of the charge Q on the plates, (iv) potential difference ΔV between the plates, and (v) energy U stored in the capacitor

Assume we apply a given potential difference ΔV_{0} to the plates.

Suppose we had not changed the geometry of the capacitor but had filled the space between the plates with a dielectric of dielectric constant, K while keeping the charge Q on the pates unchanged. How would that affect the quantities (i) - (v)?

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