A charged 3.60 g glass bead hangs suspended in the air by an electric field. If the bead was charged by the removal of 7.10×1010 electrons, what must the electric field magnitude be?  2.56×1011 N/C 3.11×106 N/C 4.97×10-13 N/C 3.17×108 N/C

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A: Electric field due to point charge(q) placed at a distance of r meter is Kq/r

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A: Given that,q1=7 nCq2=-15 nCq3=14 nCq4=-16 nC

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A: Ans:- Image-1

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A: We are allowed to give the explanation of first question only which is given below.  Please repost…

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Q: #12 Uniformly Charged Surface A thin non-metallic surface has a charge density o of 1.65 µC/m². What…

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Q: QUESTION 1 A point charge q1 = 8.0 nC is at the origin and another point charge q2 = 12.0 nC is on…

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.