Two parallel conducting plates, each of area 0.30m2, are separatedby a distance of 2.0×10−2m of air. One plate has a charge +Q;the other has a charge -Q. An electric field of 5000N/C is directedto the left in the space between the plates, as shown in thediagram. (a.) Determine the potential difference between theplates. (b.) Determine the capacitance of this arrangement ofplates. An electron is initially located at a point midway between theplates. (c.) Determine the magnitude of the electrostatic force onthe electron at this location and state its direction. (d.) If theelectron is released from rest at this location midway between theplates, determine its speed just before striking one of the plates.Assume that gravitational effects are negligible.

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Pohanginah · Accepted Answer

E=Potential Difference/d v=100 b.)C=kEoAd just plug in the numbers you have (Eo=8.85⋅10−12) k=1 c.) F=qE q=1.6⋅10−19 E=5000 d.) Calculate the EPE the electron has at rest, then convertthat into Kinetic Energy 1/2 mv2, you have the mass of an electron to be 9.1⋅10−31

Two parallel conducting plates, each of area 0.30m^2

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