The plates of a parallel plate capacitor each have an area of 0.85 m 2 and are s
ID: 2028386 • Letter: T
Question
The plates of a parallel plate capacitor each have an area of 0.85 m2 and are separated by a distance of 0.025m. They are charged until the potential difference between the plates is 3500V. The charged capacitor is then isolated.
1) How much work is required to move a -4.5C charge from the negative plate to the positive plate of this system?
2) Suppose that a dielectric sheet is inserted to completely fill the space between the plates and the potential difference between the plates drops to 2000 V. What is the capacitance of the system after the dielectric is inserted?
Explanation / Answer
1) To find how much energy is needed to move a particle, we use the formula U = qV, where U is the energy, q is the charge we are moving, and V is the change in potential the charge will experience.
U = qV
U = ( 4.5e-6 C)( 3500 V )
U = .01575 J. The field provides it with energy, meaning no external force is needed, it will move on its own. Therefore, the answer might be -.01575 J
2) We need to find the original capacitance. Since it is parallel plate, we use C = A/d.
C =( 8.85e-12 )( .85 m2 ) / ( .025 m )
C = 3 e -10 F
There is a relationship that states Cnew/Cold = Vnew / Vold
Cnew / 3 e -10 F = 3500 V / 2000 V
Cnew = 5.266 e-10 F or 52.66 nF