Question Number 163600 by Eric002 last updated on 08/Jan/22
$${a}\:{line}\:{charges}\:{of}\:{charge}\:{density}\: \\ $$$${pl}=\mathrm{4}{x}^{\mathrm{3}} −{x}+\mathrm{3}{mc}/{m}\:{laying}\:{along}\:{the}\:{x}−{axis}. \\ $$$${determine}\:{the}\:{total}\:{charge}\:{if}\:{the}\:{line}\:{charge} \\ $$$${extends}\:{from}\:{x}=\mathrm{2}\:{and}\:{x}=\mathrm{6}\:{m} \\ $$
Answered by ajfour last updated on 08/Jan/22
$$\bigtriangleup{q}=\int_{\mathrm{2}} ^{\:\mathrm{6}} \left(\mathrm{4}{x}^{\mathrm{3}} −{x}+\mathrm{3}\right){dx} \\ $$$$\:\:\:\:\:\:=\left({x}^{\mathrm{4}} −\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}\right)_{\mathrm{2}} ^{\mathrm{6}} \:\frac{{milli}\:{coulomb}}{{metre}} \\ $$$$\:\:=\mathrm{10}\left(\mathrm{36}^{\mathrm{2}} −\mathrm{18}+\mathrm{18}−\mathrm{16}+\mathrm{2}−\mathrm{6}\right)\frac{\mu{C}}{{cm}} \\ $$$$\:\:=\mathrm{10}\left(\mathrm{900}+\mathrm{360}+\mathrm{16}\right)\frac{\mu{C}}{{cm}} \\ $$$$\:\:=\mathrm{0}.\mathrm{012760}\:\frac{{C}}{{cm}}\:. \\ $$
Commented by Eric002 last updated on 08/Jan/22
$${well}\:{done}\:{sir} \\ $$