**Part a.**

The capacitance can be expressed as:

$\overline{)\begin{array}{rcl}{\mathbf{C}}& {\mathbf{=}}& \frac{\mathbf{\epsilon}\mathbf{A}}{\mathbf{d}}\\ & {\mathbf{=}}& \frac{\mathbf{k}{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}\end{array}}$, where A is the area of the parallel plates, d is the separation between the plates, k relative permeability of the dielectric material, and ε_{0} is the permeability of free space.

Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 26 nC. The area of each plate is A = 0.019 m2 The distance between them is d = 6.4. cm. The plates are in the air.

Part a. With the information given, write the equation of the capacitance in terms of ε_{0}, A, d.

Part b. Solve for the numerical value of C in pF

Part c. With the information given, express the potential difference across the capacitor in terms of Q and C.

Part d. Solve for the numerical value of ΔV in V.

Part e. If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF?

Part f. If A is then increased to 2A, d is decreased to d/4, what is the value of the capacitance C in pF?

Part g. What is the new value of potential difference ΔV in V with the original charge Q, given the values for A and d from part (f)?

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