Answer to Question #181792 in Atomic and Nuclear Physics for seulgi bae

Question #181792

A point particle of charge 2.5 nC and mass 3.25x10^-3 kg is in a uniform electric field directed to the right. It is released from rest and moves to the right. After it has traveled 12.0 cm, its speed is 25 m/s. Find the (a) work done on the particle, (b) change in the electric potential energy of the particle, and (c) magnitude of the electric field.


1
Expert's answer
2021-04-19T09:57:08-0400

To be given in question

Chage "(q)= 2.5\\times 10^{-9} =2.5nc"

mass"(m)=3.25\\times10^{-3} kg"

distance"(d)=12cm =12\\times10^{-2}meter"

Speed (v)=25 "meter\/sec"

To be asked in question

Work done (w)=?

Potential ∆U=?

Electric field (E)=?

We know that

(a)

Work energy relation r

"\u2206W=\\frac{1}{2}mv^2"

Put value

"\u2206W=\\frac{1}{2}\\times{3.25}\\times10^{-3}\\times{25}^2"

"\u2206W=1.016 jule"

(b)change in electric potential of the particle is work done on the particle to

"\u2206U=1.016 jule"

(C)

Magnitude of electric field

"F=qE"

"F.d=1.016"

"F=\\frac{1.016}{12\\times 10^{-2}} =qE"

"E=\\frac{1.016}{12\\times 10^{-2}\\times2.5\\times10^{-19}}"

"E=3.39\\times 10^ {19} N\/C"


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