Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential differen
ce is applied across the plates? (Choose from: Increasing the area of the plates; Decreasing the area of the plates; Increasing the separation between the plates; Decreasing the separation between the plates; Inserting a material with a higher dielectric constant between the plates
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
To find the scientific notation, you need to divide at the decimal by the power of 10. So since there are 2 powers of 10, what you want to do is move the decimal 2 places to the left which will give you: .054