2b) A parallel plate capacitor with a surface charge density of 9 micro-coulombs

Question

2b) A parallel plate capacitor with a surface charge density of 9 micro-coulombs/m2 has a vertical orientation and another infinite sheet with a surface charge density of +3.3 micro-coulombs/m2 lies horizontally beneath it as indicated in the sketch. If a charge of -5 micro-coulombs is to move at an angle of 27 degrees below the horizontal how far does it have to move in meters for the change in its electric potential energy to be +24 Joules?  

Explanation / Answer

electric due to charge on capacitor plates

E1x = Q/(A*eo) = (Q/A)/eo

E1x = 9*10^-6/(8.85*10^-12)

E1x = 1016949.15 N/C

electric field due to infinite sheet

E2y = sigma/(2*eo) = (3.3*10^-6)/(2*8.85*10^-12)

E2y = 186440.7 N/C

E = sqrt(E1x^2+E2y^2) = 1.034*10^6 N/C

direction = tan^-1(E2y/E1x) = 10.4

work = E*q*d*cos(27+10.4) = dU

1.034*10^6*5*10^-6*d*cos(27+10.4) = 24

distance moved d = 5.84 m

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