Answer
See answers.
Work Step by Step
a. After the dielectric is removed, the capacitance goes down by a factor of K.
b. Q=CV. The capacitance goes down by a factor of K , but the potential difference V stays the same because the capacitor is connected to a battery. So the charge decreases by a factor of K.
c. The potential difference V stays the same. It equals the battery voltage.
d. $PE=\frac{1}{2}CV^2$. The potential difference V stays the same and the capacitance decreases by a factor of K, so the energy stored also decreases by a factor of K.
e. The electric field stays the same. It equals the potential difference between the plates divided by the distance between the plates, and both quantities stay the same.
