5. A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 µC on each plate. The plates are 0.328 mm apart. a) What is the potential between the plates? b) What is the area of each plate? c) What is the electric field magnitude between the plates? d) What is the surface-charge density on each plate? e) What would the capacitance, potential difference, electric field, and surface charge density become if I replaced the air with water, which has a dielectric constant of 80, for the same charge and plate separation? difference

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5. A parallel-plate
air capacitor of
capacitance 245 pF has a charge of
magnitude 0.148 µC on each plate. The
plates are 0.328 mm apart.
a) What is the potential
between the plates?
b) What is the area of each plate?
c) What is the electric field magnitude
between the plates?
d) What is the surface-charge density on
each plate?
e)
What would the capacitance, potential
difference, electric field, and surface
charge density become if I replaced the
air with water, which has a dielectric
constant of 80, for the same charge and
plate separation?
difference
Transcribed Image Text:5. A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 µC on each plate. The plates are 0.328 mm apart. a) What is the potential between the plates? b) What is the area of each plate? c) What is the electric field magnitude between the plates? d) What is the surface-charge density on each plate? e) What would the capacitance, potential difference, electric field, and surface charge density become if I replaced the air with water, which has a dielectric constant of 80, for the same charge and plate separation? difference
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We are given capacitance of parallel plate capacitor. We are given charge on capacitor and distance between plates. We first find the potential difference of plates. We then find area. We then find the electric field magnitude between plates which is uniform.

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