Pls help me with the following question and pls pls make sure its 100% sure correct, thank you  1.  A particle with a charge of +4.20 nCnC 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 cmcm, its kinetic energy is found to be 1.50×10−6 JJ. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Parallel plates and conservation of energy. Part A.  What work was done

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Pls help me with the following question and pls pls make sure its 100% sure correct, thank you 

1. 

A particle with a charge of +4.20 nCnC 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 cmcm, its kinetic energy is found to be 1.50×10−6 JJ.

For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Parallel plates and conservation of energy.

Part A. 

What work was done by the electric force?
Express your answer with the appropriate units.
 
Part B. 
What was the change in electric potential over the distance that the charge moved?
Express your answer with the appropriate units.
 
Part C.
What is the magnitude of E⃗ �→?
Express your answer with the appropriate units.
 
Part D. 
What was the change in potential energy of the charge?
Express your answer with the appropriate units.
 
A particle with a charge of +4.20 nC is in a uniform
electric field E directed to the negative x direction.
It is released from rest, and after it has moved
8.00 cm, its kinetic energy is found to be
1.50x10-6 J.
For related problem-solving tips and strategies, you
may want to view a Video Tutor Solution of
Parallel plates and conservation of energy.
Solve: Wa→b = Kb — Ka = 1.50 × 10-6 J -0 = 1.50 × 10-6 J
Part B
What was the change in electric potential over the distance that the charge moved?
Express your answer with the appropriate units.
MÅ
AV = 4.46.10³
V
?
Units input for part B
Submit Previous Answers Request Answer
X Incorrect; Try Again; 4 attempts remaining
Transcribed Image Text:A particle with a charge of +4.20 nC is in a uniform electric field E directed to the negative x direction. It is released from rest, and after it has moved 8.00 cm, its kinetic energy is found to be 1.50x10-6 J. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Parallel plates and conservation of energy. Solve: Wa→b = Kb — Ka = 1.50 × 10-6 J -0 = 1.50 × 10-6 J Part B What was the change in electric potential over the distance that the charge moved? Express your answer with the appropriate units. MÅ AV = 4.46.10³ V ? Units input for part B Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining
A particle with a charge of +4.20 nC is in a uniform
electric field E directed to the negative x direction.
It is released from rest, and after it has moved
8.00 cm, its kinetic energy is found to be
1.50x10-6 J.
For related problem-solving tips and strategies, you
may want to view a Video Tutor Solution of
Parallel plates and conservation of energy.
What is the magnitude of E?
Express your answer with the appropriate units.
E =
Submit
Part D
O
µA
Value
Request Answer
Units
μA
?
What was the change in potential energy of the charge?
Express your answer with the appropriate units.
?
Transcribed Image Text:A particle with a charge of +4.20 nC is in a uniform electric field E directed to the negative x direction. It is released from rest, and after it has moved 8.00 cm, its kinetic energy is found to be 1.50x10-6 J. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Parallel plates and conservation of energy. What is the magnitude of E? Express your answer with the appropriate units. E = Submit Part D O µA Value Request Answer Units μA ? What was the change in potential energy of the charge? Express your answer with the appropriate units. ?
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