Question Number 34501 by rahul 19 last updated on 07/May/18 Commented by rahul 19 last updated on 07/May/18 $${ans}.\:{is}\:\:\:\:\frac{\varepsilon_{\mathrm{0}} {K}_{\mathrm{1}} {K}_{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{ln}\:\frac{{K}_{\mathrm{1}} }{{K}_{\mathrm{2}} }}{\left({K}_{\mathrm{1}}…
Question Number 34206 by rahul 19 last updated on 02/May/18 Commented by rahul 19 last updated on 04/May/18 $$? \\ $$ Terms of Service Privacy…
Question Number 34205 by rahul 19 last updated on 02/May/18 Commented by rahul 19 last updated on 02/May/18 $${ans}.\:{given}\:{is}\:\frac{\rho{x}}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\pi{a}^{\mathrm{2}} \right). \\ $$ Answered by…
Question Number 34196 by ajfour last updated on 02/May/18 Commented by ajfour last updated on 02/May/18 $$\left({i}\right){Find}\:{u}\:{if}\:{charge}\:{q}\:\left({mass}\:{m}\right) \\ $$$${is}\:{just}\:{able}\:{to}\:{cross}\:{the}\:{other}\:{ring}. \\ $$$$\left({ii}\right)\:{Can}\:{we}\:{find}\:{the}\:{time}\:{that}\:{it}\:{takes} \\ $$$${to}\:{reach}\:{from}\:{centre}\:{of}\:{lower} \\ $$$${ring}\:{to}\:{centre}\:{of}\:{upper}\:{ring},\:…
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Question Number 33912 by ajfour last updated on 27/Apr/18 Commented by ajfour last updated on 27/Apr/18 $${Find}\:{the}\:{electric}\:{field}\:{at}\:{the}\:{centre}\:{of}\: \\ $$$${a}\:{part}\:{of}\:{the}\:{surface}\:{of}\:{a}\:{spherical} \\ $$$${shell}\:{having}\:{charge}\:{density}\:+\sigma\:. \\ $$ Commented by…
Question Number 33450 by ajfour last updated on 16/Apr/18 Commented by ajfour last updated on 16/Apr/18 $${Find}\:{electric}\:{flux}\:{of}\:{a}\:{semiinfinite} \\ $$$${thread}\:{of}\:{charge}\:{having}\:{linear} \\ $$$${charge}\:{density}\:\lambda,\:{through}\:{a}\:{circular} \\ $$$${disc}\:{of}\:{radius}\:{R}\:{if}\:{one}\:{end}\:{of}\:{the} \\ $$$${thread}\:{is}\:{at}\:{the}\:{centre}\:{of}\:{this}…
Question Number 33435 by rahul 19 last updated on 16/Apr/18 $$\boldsymbol{{I}}{n}\:{a}\:{region}\:{an}\:{electric}\:{field}\:{exist}\:{in}\:{a}\:{given} \\ $$$${direction}\:{and}\:{it}\:{passes}\:{through}\:{a}\:{circle} \\ $$$${of}\:{radius}\:{R}\:{normally}.\:{The}\:{magnitude} \\ $$$${of}\:{electric}\:{field}\:{is}\:{given}\:{as}\:: \\ $$$$\boldsymbol{{E}}\:=\:{E}_{\mathrm{0}} \:\left(\mathrm{1}−\:\frac{{r}}{\boldsymbol{{R}}}\right).\:{where}\:{r}\:{is}\:{the}\:{distance} \\ $$$${from}\:{centre}\:{of}\:{circle}\:.{Find}\:{electric}\: \\ $$$${flux}\:{through}\:{plane}\:{of}\:{circle}\:{within}\:{it}. \\…
Question Number 33430 by rahul 19 last updated on 16/Apr/18 Commented by rahul 19 last updated on 16/Apr/18 $${Here}\:\boldsymbol{{E}}=\:{electric}\:{field}. \\ $$$$\boldsymbol{{V}}\:=\:{Electrostatic}\:{potential}\:. \\ $$ Commented by…
Question Number 33400 by rahul 19 last updated on 15/Apr/18 $$\boldsymbol{{Find}}\:{out}\:{electric}\:{field}\:{on}\:{an}\:{axial}\: \\ $$$${position}\:{due}\:{to}\:{a}\:{ring}\:{having}\:{linear} \\ $$$${charge}\:{density}\:\boldsymbol{\lambda}=\:\lambda_{\mathrm{0}} \:\mathrm{cos}\:\theta\:. \\ $$ Commented by ajfour last updated on 16/Apr/18…