1. The problem statement, all variables and given/known data A parallel plate capacitor has a capacitance of 1.5µF with air between the plates. The capacitor is connected a 12V battery and charged. The battery is then removed. When a dielectric is placed between the plates, a potential difference of 5.0V is measured across the plates. What is the dielectric constant of the material? 2. Relevant equations not sure on this, possible culprits are: q=CV where as far as I understand q is magnitutde of charge on each plate c is capacitance and v is voltage k=E0/E This is the only formula in our notes that doesnt involve plate area and distance between plates, which isn't given in the question. What E0 and E are isn't explained though which is causing me no end of confusion Is it fair to say V=12 in this question and C=1.5(10^-6)? 3. The attempt at an answer I tried googling the question and funilly enough it came up here before. The OP was equally clueless and the advice given was: a dielectric will increase the capacitance but charge on the the plates will be same. Can you write an equation with this information and find the new capacitance? Comparing the two capacitances, you can calculate the dielectric constant of the new capacitor. It's late at night so can someone please just tell me what the E0/E formula means and what formulae are relevant to the question? I asked around but the question is "sorcery" to everyone I know. Is the formula I attached below relevant? It is dependant on both capacitance and dialectric constant which makes it seem relevant but it invovles variables not given in the question (distance and area).