The cylindrical imaging drum of a photocopier is to have an electric field just

Question

The cylindrical imaging drum of a photocopier is to have an electric field just outside its surface of 3.00 105 N/C. (a) If the drum has a surface area of 0.0610 m^2 (the area of a 8 1/2 multiplied by 11 in. sheet of paper), what total quantity of charge must reside on the surface of the drum? (b) If the surface area of the drum is increased to 0.122 m^2 so that larger sheets of paper can be used, what total quantity of charge is required to produce the same 3.00*10^5 N/C electric field just above the surface?

Explanation / Answer

total electric flux = charge(Q) / eo E * surface area = charge(Q) / eo Q = e0 * 3×10^5 *0.0610m2 hope area and E are in MKS Q = 8.85×10-12 * 3×10^5 *0.0610 = 1.61955*10^-07

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