Question Number 46293 by Umar last updated on 23/Oct/18 Commented by Umar last updated on 23/Oct/18 $${please}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 46292 by Umar last updated on 23/Oct/18 $${A}\:+{ve}\:{point}\:{charge}\:{of}\:{magnitude}\:{q}\:{is}\:{located}\:{on}\:{the}\:{y}−{axis}\: \\ $$$${at}\:{a}\:{point}\:{of}\:{y}=+{d}\:{and}\:{another}\:−{ve}\:\:{charge}\:{of}\:{same}\:{magnitude} \\ $$$${is}\:{located}\:{on}\:{the}\:{point}\:{y}=−{d}.\: \\ $$$${A}\:{third}\:+{ve}\:{charge}\:{of}\:{same}\:{magnitude}\:{is}\:{located}\: \\ $$$${at}\:{some}\:{point}\:{on}\:{x}−{axis}. \\ $$$$\:\:\:\left(\mathrm{1}\right)−\:{what}\:{is}\:{the}\:{magnitude}\:{and}\:{direction}\:{of}\:{the}\:{force}\:{on} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{third}\:{charge}\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}\right)−\:{when}\:{its}\:{located}\:{on}\:{the}\:{origin}. \\…
Question Number 107764 by hhryhrry2 last updated on 12/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\theta\:=\:\:\frac{\Sigma\:\mathrm{F}_{\mathrm{y}} }{\Sigma\:\mathrm{F}_{\mathrm{x}} } \\ $$$$ \\ $$ Commented by ajfour last updated on 12/Aug/20 $$\frac{{d}\bar {{p}}}{{dt}}=\sqrt{\left(\Sigma{F}_{{x}}…
Question Number 41530 by physics last updated on 09/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18 $${E}=\frac{\lambda}{\mathrm{2}\Pi\epsilon_{\mathrm{0}} {r}_{\bot} }\:\:{E}={electric}\:{field}\:{at}\:{point}\:{which}\:{is}\:{ag} \\ $$$${a}\:{perpedicular}\:{distance}\:{r}_{\bot} \:{from}\:{infinite}\:{charge}\:{rod}= \\ $$$$\lambda={charge}\:{per}\:{unit}\:{length} \\…
Question Number 172286 by cyathokoza last updated on 25/Jun/22 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 40817 by Necxx last updated on 27/Jul/18 $${An}\:{electric}\:{field}\:{of}\:\mathrm{2}.\mathrm{66}{KV}/{m},{and} \\ $$$${a}\:{magnetic}\:{field}\:{of}\:\mathrm{1}.\mathrm{20}{T},{act}\:{on}\:{a} \\ $$$${moving}\:{electron}.{Calculate}\:{the}\:{speed} \\ $$$${of}\:{the}\:{electron}\:{if}\:{the}\:{resultant}\:{force} \\ $$$${due}\:{to}\:{both}\:{fields}\:{is}\:{zero}. \\ $$$$ \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 40816 by Necxx last updated on 27/Jul/18 $${An}\:{electric}\:{field}\:{of}\:\mathrm{2}.\mathrm{66}{KV}/{m},{and} \\ $$$${a}\:{magnetic}\:{field}\:{of}\:\mathrm{1}.\mathrm{20}{T},{act}\:{on}\:{a} \\ $$$${moving}\:{electron}.{Calculate}\:{the}\:{speed} \\ $$$${of}\:{the}\:{electron}\:{if}\:{the}\:{resultant}\:{force} \\ $$$${due}\:{to}\:{both}\:{fields}\:{is}\:{zero}. \\ $$$$ \\ $$ Terms of Service…
Question Number 40804 by rishabh last updated on 27/Jul/18 Commented by rishabh last updated on 27/Jul/18 $$\mathrm{My}\:\mathrm{question}\:\mathrm{is}\:\mathrm{which}\:\mathrm{approach} \\ $$$$\mathrm{is}\:\mathrm{correct}\:\mathrm{and}\:\mathrm{why}? \\ $$$$\mathrm{book}\:\mathrm{gives}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{approach}\:\mathrm{2}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{using}\:\mathrm{approach} \\ $$$$\mathrm{1}.…
Question Number 40786 by Necxx last updated on 27/Jul/18 $${A}\:{parallel}\:{plate}\:{capacitor}\:{of}\:{plate} \\ $$$${spacing},\:\mathrm{1}{mm}\:{is}\:{charged}\:{to}\:{a}\:{potential} \\ $$$${of}\:\mathrm{50}{V}.{Find}\:{the}\:{energy}\:{density}\:{in} \\ $$$${the}\:{capacitor} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18…
Question Number 40773 by Necxx last updated on 27/Jul/18 $${A}\:{pair}\:{of}\:{oppositely}\:{charged}\:{plane} \\ $$$${parallel}\:{plates}\:{each}\:{of}\:{area},\mathrm{100}{cm}^{\mathrm{2}} \\ $$$${has}\:{the}\:{electric}\:{field}\:{of}\:{the}\:{value} \\ $$$$\mathrm{5}.\mathrm{0}×\mathrm{10}^{\mathrm{4}} {N}/{C}.{Calculate}\:{the}\:{charge} \\ $$$${on}\:{each}\:{plate}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last…