A parallel plate capacitor is given a charge Q. After charging, the capacitor is
ID: 1873344 • Letter: A
Question
A parallel plate capacitor is given a charge Q. After charging, the capacitor is removed from the circuit so that the charge remains fixed. The potential energy stored on the capacitor is A student inserts a dielectric into the space between the plates, effectively reducing the electric field between the plates by a factor equal to the dielectric constant K. This increases the capacitance by a factor K. QUESTION 1. Has the electrical energy stored in the capacitor increased or decreased? Defend your answer. QUESTION 2 2. Did the student do positive or negative work on the dielectric? Again, defend your answer. Hint: Be sure the two responses are consistent with one another.Explanation / Answer
1) energy = q2 /2C
since capacitance C will increase due to dielectric so energy will decrease.
2) the side of dielectric closest to positive capacitor plate is negatively charged and the side closest to negative capacitor plate is positively charged.so the dielectric is attracted to capacitor. hence student do negative work.