Help me!! A parallel plate capacitor of capacitance C is constructed from two th
ID: 2143180 • Letter: H
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Help me!!
A parallel plate capacitor of capacitance C is constructed from two thin conducting metal plates each having area A - 1.50 cm2. They are separated by a dielectric spacer d = 2.00 mm thick as shown in Fig. P3. Neglecting the effects of fringing fields, determine the following: (4) The value of the dielectric constant k of the spacer layer if the capacitance C is 2.00x10-12F. A single 6.00 V battery is now connected across the capacitor. (5) Compute the surface charge density sigma of the negatively-charged plate. The battery is now disconnected from the capacitor. Determine: (4) The energy U stored by the capacitor. (5) The magnitude of the electric field (E) between the two plates. The plates are squeezed together so that they are now spaced by d' - 1.00 mm. Assume the dielectric constant of the spacer layer remains unchanged. Now compute: (5) The new potential difference V across the two plates. (2) The work W done by the external agent that squeezed the plates closer. Figure P3 A parallel plate capacitor. The conducting plates are on top (shown) and bottom (not shown).Explanation / Answer
Too many questions less points so I am just writing the formulas please solve for yourself
d is distance between plates, C is capacitance, e_0 = 8.85 * 10^-12 and A is the area, V is the voltage applied
a) k = (C*d)/(e_0 * A)
b) Charge = C*V = CV
Charge density = CV/A
c) Energy stored in capacitor = 0.5CV^2 = 0.5CV^2
d) Magnitude of electric Field = CV/(A*e_0)
e) Charge remains same when we squeeze the capacitor
C' = Cd/d' , where d' is the new distance between the plates:
Charge is same so CV
new Potential V' = CVd'/Cd = Vd'/d
f) Work done by external force = 0.5C'V'^2 - 0.5CV^2
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