A 7.5 g glass bead is charged by the removal of 2.5 x 10^10 electrons. What electric field strength (in units of N/C) will cause the bead to hang suspended in the air? N/C
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A 7.5 g glass bead is charged by the removal of 2.5 x 10^10 electrons. What electric field strength (in units of N/C) will cause the bead to hang suspended in the air?
N/C
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- 2A small plastic bead has a mass of 4.02 g and a charge of -17.4 µC. It levitates, motionless, when placed in a uniform electric field perpendicular to the ground. What is the magnitude of the electric field (in N/C)? N/C What is the direction of the electric field? O upward O downward O east O westA 148 mg plastic bead has a radius of 344 μm and is charged by the addition of 1.38×1010 excess electrons. You may assume that the excess electrons are spread uniformly throughout the bead. What electric field strength will cause the bead to hang suspended in the air?
- Two tiny particles having charges +10.0 nC and -10.0 nC are separated by a distance of 20.0 cm. What are the magnitude and direction of electric field midway between these two charges? O 9.0 x 10^3 N/C directed towards the negative charge O 9.0 x 10^3 N/C directed towards the positive charge O The net electric field is zero. O 2.3 x 10^3 N/C directed towards the positive charge O 2.3 x 10^3 N/C directed towards the negative chargeDetermine the magnitude of the electric field.Two plastic balls, each of radius 17.9 cm, are placed so that there is a 36.6 cm space between them. The charge on one ball is +3.20 nC and the charge on the other is -4.70 nC. What is the magnitude of the electric field at a point directly midway between the two balls? 2.22×103 N/C 542 N/C 103 N/C 530. N/C
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