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)**

Using the expression for the capacitance:

C_{0} = ε_{0}A/l

C = ε_{0}A/2l

C/C_{0} = (ε_{0}A/2l)/(ε_{0}A/l) = 1/2

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 double the plate separation while keeping the potential difference constant, calculate the quantities (i) - (v), for this case, and express each in terms of the original value, e.g., C = 2C_{0}, E = (1/4)E_{0}, etc. (Obviously, ΔV = ΔV_{0}).

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