Question and Answers Forum

All Questions      Topic List

Electrostatics Questions

Previous in All Question      Next in All Question      

Previous in Electrostatics      Next in Electrostatics      

Question Number 170297 by ali009 last updated on 19/May/22

 find the total charge enclosed inside a   volume of 1.5×10^(−9)  located at the origin  if D=e^(−x) sin(y)a_x ^� −(e^(−x) cos(y))a_y ^� +2za^� z C/m^2

$$\:{find}\:{the}\:{total}\:{charge}\:{enclosed}\:{inside}\:{a}\: \\ $$$${volume}\:{of}\:\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{9}} \:{located}\:{at}\:{the}\:{origin} \\ $$$${if}\:{D}={e}^{−{x}} {sin}\left({y}\right)\hat {{a}}_{{x}} −\left({e}^{−{x}} {cos}\left({y}\right)\right)\hat {{a}}_{{y}} +\mathrm{2}{z}\hat {{a}z}\:{C}/{m}^{\mathrm{2}} \\ $$

Answered by ali009 last updated on 20/May/22

pv=▽.D^→   pv=−e^(−x) sin(y)+e^(−x) sin(y)+2=2  Q_(enc) =∫_v pv . dv  Q_(enc) =pv∫dv  Q_(enc) =pv . v=1.5×10^(−9) ×2=3 nC

$${pv}=\bigtriangledown.\overset{\rightarrow} {{D}} \\ $$$${pv}=−{e}^{−{x}} {sin}\left({y}\right)+{e}^{−{x}} {sin}\left({y}\right)+\mathrm{2}=\mathrm{2} \\ $$$${Q}_{{enc}} =\int_{{v}} {pv}\:.\:{dv} \\ $$$${Q}_{{enc}} ={pv}\int{dv} \\ $$$${Q}_{{enc}} ={pv}\:.\:{v}=\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{9}} ×\mathrm{2}=\mathrm{3}\:{nC} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com