Question Number 170297 by ali009 last updated on 19/May/22 $$\:{find}\:{the}\:{total}\:{charge}\:{enclosed}\:{inside}\:{a}\: \\ $$$${volume}\:{of}\:\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{9}} \:{located}\:{at}\:{the}\:{origin} \\ $$$${if}\:{D}={e}^{−{x}} {sin}\left({y}\right)\hat {{a}}_{{x}} −\left({e}^{−{x}} {cos}\left({y}\right)\right)\hat {{a}}_{{y}} +\mathrm{2}{z}\hat {{a}z}\:{C}/{m}^{\mathrm{2}} \\ $$ Answered…
Question Number 104572 by ajfour last updated on 22/Jul/20 Commented by ajfour last updated on 22/Jul/20 $${A}\:{point}\:{charge}\:{q}\:{with}\:{mass}\:{m} \\ $$$${is}\:{released}\:{at}\:{a}\:{distance}\:\boldsymbol{{a}}\:{from} \\ $$$${a}\:{fixed}\:{uniformly}\:{charged}\:{rod} \\ $$$${of}\:{length}\:{L},\:{charge}\:{Q}.\:{Find}\:{the} \\ $$$${speed}\:{acquired}\:{by}\:{point}\:{charge}…
Question Number 38675 by rahul 19 last updated on 28/Jun/18 Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18 $${A}\rightarrow{Q},{B}\rightarrow{Q} \\ $$ Commented by rahul 19 last…
Question Number 38291 by Tinkutara last updated on 23/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18 $${E}=\frac{\mathrm{1}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} }×\frac{{e}}{{d}^{\mathrm{2}} }\:\:\:{e}={charge}\:{of}\:{electron} \\ $$$$=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{1}×\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} }{\left(\mathrm{10}^{−\mathrm{10}} \right)^{\mathrm{2}} }=\mathrm{14}.\mathrm{4}×\mathrm{10}^{\mathrm{9}−\mathrm{19}+\mathrm{20}}…
Question Number 37649 by Tinkutara last updated on 16/Jun/18 Answered by Rio Mike last updated on 16/Jun/18 $$\mathrm{yes}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\left(\mathrm{C}\right)\:\mathrm{net}\:\mathrm{induced} \\ $$$$\mathrm{charge}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{is}\:\mathrm{zero}.\mathrm{If}\:\mathrm{charges} \\ $$$$\mathrm{can}\:\mathrm{neither}\:\mathrm{flow}\:\mathrm{onto}\:\mathrm{or}\:\mathrm{off}\:\mathrm{the}\: \\ $$$$\mathrm{spherical}\:\mathrm{shell}\:\mathrm{the}\:\mathrm{net}\:\mathrm{charge}\:\mathrm{cannot} \\…
Question Number 37058 by Tinkutara last updated on 08/Jun/18 Answered by ajfour last updated on 08/Jun/18 $${E}_{{cube}} +{E}_{{q}} =\mathrm{0} \\ $$$${E}_{{cube}} =−{E}_{{q}} =\frac{−{q}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {d}^{\:\mathrm{2}} }\:.…
Question Number 37045 by ajfour last updated on 08/Jun/18 Commented by ajfour last updated on 08/Jun/18 $${Find}\:{electric}\:{field}\:{at}\:{a}\:{distance} \\ $$$${r}_{\bot} \:{from}\:{a}\:{line}\:{charge}\:{of}\: \\ $$$${uniform}\:{charge}\:{density},\:{net} \\ $$$${charge}\:{q}. \\…
Question Number 37032 by ajfour last updated on 08/Jun/18 Commented by ajfour last updated on 08/Jun/18 $${Find}\:{Electric}\:{field}\:{at}\:{P}\:\:{due}\:{to} \\ $$$${the}\:{charge}\:{contained}\:{in}\:{the}\: \\ $$$${paraboloid}\:{having}\:{a}\:{uniform} \\ $$$${volume}\:{charge}\:{density}\:\boldsymbol{\rho}. \\ $$…
Question Number 36980 by Tinkutara last updated on 07/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18 $${charge}\:{densigy}=\lambda \\ $$$${let}\:{length}\:{of}\:{rod}={l}=\mathrm{10}{cm} \\ $$$${distance}\:{of}\:{point}\:{p}\:{from}\:{the}\:{centre}\:{of}\:{rod}\:{is} \\ $$$${h}=\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}{l}….{as}\:{per}\:{question}\:{point}\:{p}\:{and}\:{two}\: \\ $$$${end}\:{of}\:{rods}\:{make}\:{a}\:{equilateral}\:\:{triangle}…
Question Number 36852 by Tinkutara last updated on 06/Jun/18 Terms of Service Privacy Policy Contact: info@tinkutara.com