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Question Number 31740 by Cheyboy last updated on 13/Mar/18

Consider a particle in a uniform  charge electric field.If the  particle has a charge of  2C  an is placed 3m away from  the charge flate.  Calculate the work needed  to move the 2C particle  to a distance of 1m from the  plate.    Plzz help

$$\boldsymbol{\mathcal{C}{onsider}}\:\boldsymbol{{a}}\:\boldsymbol{{particle}}\:\boldsymbol{{in}}\:\boldsymbol{{a}}\:\boldsymbol{{uniform}} \\ $$$$\boldsymbol{{charge}}\:\boldsymbol{{electric}}\:\boldsymbol{{field}}.\boldsymbol{{If}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{particle}}\:\boldsymbol{{has}}\:\boldsymbol{{a}}\:\boldsymbol{{charge}}\:\boldsymbol{{of}}\:\:\mathrm{2}\boldsymbol{{C}} \\ $$$$\boldsymbol{{an}}\:\boldsymbol{{is}}\:\boldsymbol{{placed}}\:\mathrm{3}\boldsymbol{{m}}\:\boldsymbol{{away}}\:\boldsymbol{{from}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{charge}}\:\boldsymbol{{flate}}. \\ $$$$\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{work}}\:\boldsymbol{{needed}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{move}}\:\boldsymbol{{the}}\:\mathrm{2}\boldsymbol{{C}}\:\boldsymbol{{particle}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{a}}\:\boldsymbol{{distance}}\:\boldsymbol{{of}}\:\mathrm{1}\boldsymbol{{m}}\:\boldsymbol{{from}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{plate}}. \\ $$$$ \\ $$$$\boldsymbol{{Plzz}}\:\boldsymbol{{help}} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18

Lorentz force=qE    q=charge of charge particle    E=eletric field strength for charge plate    work done=qEd  d=displacement  q=2C   d=1mm  E=(σ/(2ε_0 ))(for nonconducting sheet)  E=(σ/ε_0 )  (for conducting sheet)  in question σ=not given  in question sheet type whether conducting  or non conducting not mentioned    (W)_c =q×(σ/ε_0 )×d  for conducting sheet  (W)_(NC) =q×(σ/(2ε_0 ))×d  put the value of q=2C    d=1mm

$${Lorentz}\:{force}={qE}\:\:\:\:{q}={charge}\:{of}\:{charge}\:{particle} \\ $$$$\:\:{E}={eletric}\:{field}\:{strength}\:{for}\:{charge}\:{plate} \\ $$$$ \\ $$$${work}\:{done}={qEd} \\ $$$${d}={displacement} \\ $$$${q}=\mathrm{2}{C}\:\:\:{d}=\mathrm{1}{mm} \\ $$$${E}=\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }\left({for}\:{nonconducting}\:{sheet}\right) \\ $$$${E}=\frac{\sigma}{\epsilon_{\mathrm{0}} }\:\:\left({for}\:{conducting}\:{sheet}\right) \\ $$$${in}\:{question}\:\sigma={not}\:{given} \\ $$$${in}\:{question}\:{sheet}\:{type}\:{whether}\:{conducting} \\ $$$${or}\:{non}\:{conducting}\:{not}\:{mentioned} \\ $$$$\:\:\left({W}\right)_{{c}} ={q}×\frac{\sigma}{\epsilon_{\mathrm{0}} }×{d}\:\:{for}\:{conducting}\:{sheet} \\ $$$$\left({W}\right)_{{NC}} ={q}×\frac{\sigma}{\mathrm{2}\epsilon_{\mathrm{0}} }×{d} \\ $$$${put}\:{the}\:{value}\:{of}\:{q}=\mathrm{2}{C}\:\:\:\:{d}=\mathrm{1}{mm} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 24/May/18

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