Question

A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. If, instead of separating the plates, the empty space between the plates is filled with a slab of insulating material that has a dielectric constant κ= 2, which of the following correctly describes the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor? 1) Voltage halvesElectric field halves Energy halves 2. Voltage halves Electric field halves Energy does not change 3. Voltage does not change Electric field does not change Energy halves 4. Voltage does not changeElectric field halves Energy halves 5. Voltage does not change Electric field does not change Energy does not change

Answer 1

The capacitance of a capacitor of a parallel plate is given by:

$C=\frac{kA\cdot\epsilon_o}{d}$

Where:

$\begin{gathered} \epsilon_o=_{\text{ }}Vacuum_{\text{ }}permittivity \\ k=_{\text{ }}Dielectric_{\text{ }}constant=2 \\ A=_{\text{ }}Area_{\text{ }}of_{\text{ }}the_{\text{ }}plates \\ d=_{\text{ }}distance_{\text{ }}betwen_{\text{ }}the_{\text{ }}plates \end{gathered}$

We also know:

$\begin{gathered} Q=\frac{q}{V} \\ so\colon \\ \frac{q}{V}=\frac{kA\cdot\epsilon_o}{d} \\ V=\frac{qd}{kA\cdot\epsilon_o} \\ if \\ k=2 \\ V=\frac{1}{2}(\frac{qd}{kA\cdot\epsilon_o}) \end{gathered}$

As we can see the voltage halves

And since:

$\begin{gathered} E=\frac{V}{d} \\ W=qV \end{gathered}$

We can conclude that electric field and also the energy halve