I have not been able to get the first part, the answer to teh second part would

Question

I have not been able to get the first part, the answer to teh second part would be helpful too. Problem 21.48 A -60nC charged particle is in a uniform electric field E= (20V/m , east). An external force moves the particle 1.0m north, then 6.0m east, then 5.0m south, and finally 3.0m west. The particle begins and ends its motion with zero velocity. Part A How much work is done on it by the external force? Express your answer to two significant figures and include the appropriate units. W=3.6.10^-5 Part B What is the potential difference between the particle's final and initial positions? Express your answer to two significant figures and include the appropriate units. Delta V =Value Units

Explanation / Answer

Here ,

let us assume east as +x and north as +y direction

Then

displacement = 1j +6i - 5j -3i

d = 3i - 4 j

Now , E = 20 i V/m

as Work done by external force W = -q(E.d)

W = 50 * (3i - 4 j ).(20)

W = 3000 nJ = 3*10^-6 J

the work done by external force is 3*10^-6 J

B)

Here , Potential difference = Work done/charge

Potential difference = 3000/(-50)

Potential difference = -60 V

the potential difference is -60 V

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