(a) Addnegative charge to the metal slab. (b) Addpositive charge to the metal sl
ID: 1730114 • Letter: #
Question
(a) Addnegative charge to the metal slab.
(b) Addpositive charge to the metal slab.
(c) Changing the charge on the metal slab willnot affect the potentialdifference Vab.
The answer is b but if someone could explain why it would bemost helpful. Thanks
Explanation / Answer
a < 0, the electric field due toa is to the left, b > 0, the electric field due tob is to the left, the net field due to a and bis E = (2.5 + 7.5)*10-6/(20) =5.647*105 V/m the potential difference due to aand b = Vb' - Va' =5.647*105 V/m * 0.12 m = 67.8 kV the question gives |Vab| = 50 kV, butdoesn't say 50 kV = Vb - Va,or Va - Vbif Vb -Va = 50 kV < 67.8 kV, the field between 2 insulatorplates due to the metal plate should be to the right, sometal must be positive. if Va - Vb = 50 kV, the field between 2insulator plates due to the metal plate should be to the rightand larger than E, so metal must bepositive. So in either case, the answer is b)Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.