A particle has charge 2.10 nC. (a) Find the magnitude and direction of the field

Question

A particle has charge 2.10 nC. (a) Find the magnitude and direction of the field due to this particle at a point 0.220 m directly above it. magnitude. N/C (b) At what distance from this particle doss Its electric field have a magnitude of 10.0 N/C? A uniform electric field exists in the region between two oppositely charged plane parallel plates A proton is released from rest at the surface of the positively changed plate and strikes the strikes to the opposite plate 2.20 cm distant from the first, in a time interval of 1.00 times 10^-6 z (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate. Three negative point charges lie along a lime as shown in the following figure (d_2 = 8.20 cm, and a = 5.10 mu c). Find the magnitude and direction of the electric field measured perpendicular to the line connecting the three charges. (Take the +x direction to be to the right.)

Explanation / Answer

Part a

Charge on given particle Q=2.10 nC

Magnitude of electric field at a distance r fron it is

                E=KQ/r^2

Where K =9*10^9 N-m^2/C^2

At a distance r=0.220 m

                E=9* 10^9 * 2.10* 10^-9 /(0.220)^2

                E=390.4959 N/C

Field will be in upward direction.

Part b

Let at a distance d field of charge is E=10 N/C.

Then          E=KQ/d^2

                d^2=KQ/E

                d^2 = 9*10^9 *2.10 *10^-9 / 10

                d^2 = 1.89

                d=1.3747 m

  

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