Answer:
increasing the separation between the plates
Explanation:
The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because
Q(charge) = CV V = VOLTAGE , c = capacitance
E = 1/2 eAV^2/ D ( ENERGY STORED )
where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference
5.6 g/ml. That is the density.
D. is the right answer because his pressure is very bad out there in the air.
good luck
Answer:
the answer is at the BOTTOM OF THEIR QUESTION
Explanation:
IT IS CORRECT BTW
