A particle with a charge of +4.30 nC is in a uniform electric field E directed t

Question

A particle with a charge of +4.30 nC is in a uniform electric field E  directed to the negative x direction. It is released from rest, and after it has moved 8.00cm , its kinetic energy is found to be 1.50×106 J .

Part A

What work was done by the electric force?

Express your answer in joules to three significant figures.

1.50×106

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Part B

What was the change in electric potential over the distance that the charge moved?

Express your answer in volts to three significant figures.

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Part C

What is the magnitude of E ?

Express your answer in volts per meter to three significant figures.

4360

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Part D

What was the change in potential energy of the charge?

Express your answer in volts per meter to three significant figures.

A particle with a charge of +4.30 nC is in a uniform electric field E  directed to the negative x direction. It is released from rest, and after it has moved 8.00cm , its kinetic energy is found to be 1.50×106 J .

Part A

What work was done by the electric force?

Express your answer in joules to three significant figures.

W =

1.50×106

  J  

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Correct

Part B

What was the change in electric potential over the distance that the charge moved?

Express your answer in volts to three significant figures.

V= V

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Incorrect; Try Again

Part C

What is the magnitude of E ?

Express your answer in volts per meter to three significant figures.

E =

4360

  V/m  

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Correct

Part D

What was the change in potential energy of the charge?

Express your answer in volts per meter to three significant figures.

U= J

Explanation / Answer

Here

charge ,q = 4.3 nC

distance ,d = 8 cm = 0.08 m

kinetic energy , KE = 1.5 *10^-6 J

a) Using work energy theorum

work done by electric field = Kinetic energy

work done by electric field = 1.5 *10^-6 J

b)

change in electric potential = -workdone by electric field/q

change in electric potential = -1.5 *10^-6/(4.3 *10^-9)

change in electric potential = - 348.9 V

the change in electric potential is - 348.9 V

c)

for the electric field

electric field = V/d

electric field = 348/.08 N/C

electric field = 4360 N/C

d)

change in potential energy = - change in kinetic energy

change in potential energy = -1.5 *10^-6 J

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