A uniform electric field of magnitude 410N/C pointing in the positive x -directi

Question

A uniform electric field of magnitude 410N/C pointing in the positive x-direction acts on anelectron, which is initially at rest. The electron has moved3.40 cm. (a) What is the work done by the field on theelectron?
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(b) What is the change in potential energy associated with theelectron?
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(c) What is the velocity of the electron?
magnitude m/s direction ---Select----x+y-y+x (a) What is the work done by the field on theelectron?
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(b) What is the change in potential energy associated with theelectron?
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(c) What is the velocity of the electron?
magnitude m/s direction ---Select----x+y-y+x magnitude m/s direction ---Select----x+y-y+x

Explanation / Answer

   a)The work done by the field o n theelectron is                  W = (e E ) d Here e =1.6*10-19C            E = 410 N/C            d = 0.034m           W =(1.6*10-19C)(410 N/C)( 0.034m) b) The change in potential energy with theelectron is              V = W                   =  (e E ) d                    =  (1.6*10-19C)(410N/C)( 0.034m) c)   According to law of coservation ofenergy       W = Change in kineticenergy            =( 1/2) mv2 -0 The speed of the electron is             v = [W /2m] Substitute the values. Direction is +x            

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