Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is dį = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of V₂ = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vf between the plates after the plates have been pulled to their new, larger separation df. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will need to do work. We expect that this work will be positive- why is that? Provide a justification. Hint: You could consider the change in the stored energy, and/or the force exerted by the plates on each other and the force we have to exert. Then use your results from part (c) to calculate the amount of work you will need to do to separate the plates, and check that the work is indeed positive.

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To calculate the new potential difference and the electrostatic potential energy stored in the capacitor before and after.

Given:
Plate area: $A=8.50 {\text{cm}}^2=8.50\times {10}^{-4} {\text{m}}^2$
Initial separation: $d_i=3.00 \text{mm}=3.00\times {10}^{-3} \text{m}$
Initial potential difference: $V_i=6.00 \text{V}$
Final separation: $d_f=8.00 \text{mm}=8.00\times {10}^{-3} \text{m}$
Vacuum permittivity: ${\epsilon }_0=8.85\times {10}^{-12} \text{F/m}$
Dielectric constant of air: $k_{air}=1$

To calculate:
(a) Relation between the new potential difference and the old potential difference.
(b) The new potential difference: $V_f$
(c) Initial Electrostatic energy stored in the capacitor: $U_i$
     Final Electrostatic potential energy stored in the capacitor: $U_f$

The capacitance of a parallel plate capacitor is given by:
$C=k{\epsilon }_0\frac{A}{d}$
Where,
$k$ is the dielectric constant,
${\epsilon }_0$ is the Vacuum Permittivity,
$A$ is the area of the plates and
$d$ is the separation of the plates.

The charge on a capacitor is given by the relation: $Q=CV$
Where, $C$ is the capacitance and $V$ is the voltage.

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