Question

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Problem 2: (10 points) A-5PC Test charge is placed at the point P(2,2,0) near a plane with p. = 1.5nC/m², the plane is defined as z = 4, a) Calculate the force that the plane will exert on the test charge. b) How would the magnitude and direction of the force change if the magnitude of the test charge was decreased by half and its sign changed with everything else remaining the same?

Answer 1

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- 9. A charge of suc Bplaced at c2,2,0) & a surface at ze z=4 with Ss=1.5nelm? then force exerted on charge=? → first calculating Electric field intensity at (2,2,0) due to surface charge at 3=4 16,3,0) Ē=5s år where ân = Unit normal - Vector 1 to surface - >a 1. ån=tve if point of inte 이 R =4 is above surface s ân=-ve if p.o I is below surface here (2,2,0) de 2=0 s below 2=4 äne-az - where εo = 10 azo 'Is=150/m² 36T - & Neistos Er is PĒ= 180x1.5. The cry -F- -2717 a) cym XI I force exerted on point charge by Istanc/m2 B F= -27.1. 22x-5J4C E = 4.246zx104 NI

b) Now 2. Thus Q = 2.5oC ( magnitude = { f sign changed). F = (2.5mor).(-271 âz) Fa-2.12@2x1074 Newton Hence as we can see Thos with sage Fenew) = fold) of due to sign change in polarity of charge direction of force Reversed