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A particle with a charge of +4.20 nC is in a uniform electric field E⃗ directed to the negative x direction. It is released from rest, and after it has moved 8.00 cm , its kinetic energy is found to be 1.50×10−6 J .A. What work was done by the electric force?B. What was the change in electric potential over the distance that the charge moved?C. What was the magnitude of E? in V/MD. What was the change in potential energy of the charge? in J

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Answer:

a. 1.50×10⁻⁶ J

b.  357.1 V

c. 4 463 N/C

d. zero

Explanation:

We are given:

q= +4.20 nC

 = + 4.20 × 10 ⁻⁹ C

The charge moves from rest to a point in the negative direction. Therefore, the  kinetic energy of the charge at the point of origin is Ka = 0

the distance, l = 8 cm

                        = 0.08 m

a. Work done by electric field = Ka

                                                 = 1.50×10⁻⁶ J

b. the change in electric potential = [tex]\frac{V}{q}[/tex]

                                                         = [tex]\frac{1.50 x 10-6}{4.20x10-9}[/tex]

                                                         = 357.1 V

c. magnitude                                  = [tex]\frac{357.1}{0.08}[/tex]

                                                       = 4 463 N/C

d. zero

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