Problem 21.48 A -50nC charged particle is in a uniform electric field vector E =

Question

Problem 21.48 A -50nC charged particle is in a uniform electric field vector E = (30V/m , east). An external force moves the particle 4.0m north, then 6.0m east, then 7.0m south, and finally 4.0m 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 = 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 =

Explanation / Answer

Here , let be +x - axis and north be +y - axis

E = 30 V/m i

q = -50 nC

Now , displacement , d = 4 j + 6 i - 7 j - 4 i

d = 2i -3j

Now , Work done = -q*(E.d)

Work done = 50 * (2i -3j ).(30i)

Work done = 3000 nJ

work done by external force is 3000 nJ

B)

Here , Potential difference = work done by field/ charge

Potential difference = 3000/-50

Potential difference = -60 V

the potential difference is -60 V

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