Question Number 52222 by Umar last updated on 04/Jan/19 Answered by peter frank last updated on 04/Jan/19 $$\left.{a}\right)\:\:\:{E}=\frac{{v}}{{d}} \\ $$$${E}=\frac{\mathrm{150}}{\mathrm{6}×\mathrm{10}^{−\mathrm{3}} }=\mathrm{25000}{V}/{m} \\ $$$${a}=\frac{{eE}}{{m}} \\ $$$${a}=\frac{\left(\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}}…
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Question Number 50994 by peter frank last updated on 23/Dec/18 $${Two}\:{similar}\:{ball}\:{of}\:{mass} \\ $$$${m}\:{attached}\:{by}\:\:{silk}\:{thread} \\ $$$${of}\:{length}\:{a}\:\:{and}\:{carry} \\ $$$${similar}\:{charge}\:\:{q}.{assume}\:\theta\:{is} \\ $$$${small}\:{enough}\:{that} \\ $$$${tan}\theta\approx{sin}\theta\:{to}\:{this} \\ $$$${approximation},{show}\: \\ $$$${that}\:\:\:\:{X}=\left(\frac{{qa}}{\mathrm{2}\pi\varepsilon_{\mathrm{0}}…
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Question Number 48458 by ajfour last updated on 24/Nov/18 Commented by ajfour last updated on 24/Nov/18 $${Q}.\mathrm{48405}\:\:\:\:\:\left({attempting}\:{to}\:{solve}\right) \\ $$ Answered by ajfour last updated on…
Question Number 48405 by ajfour last updated on 23/Nov/18 Commented by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18 $${now}\:{pls}\:{explain}\:… \\ $$$${when}\:{the}\:{ring}\:{is}\:{charged}\:{by}\:{charge}\:{Q}, \\ $$$${then}\:{why}\:{tension}\:{originate}… \\ $$$${what}\:{is}\:{the}\:{cause}\:{of}\:{origin}\:{of}\:{tension}… \\ $$$${Tension}\:{originate}\:{if}\:{you}\:{place}\:{another}…
Question Number 47646 by Umar last updated on 12/Nov/18 Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18 Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18 Answered by…
Question Number 47641 by azharkhan250963@gmail.com last updated on 12/Nov/18 $$\mathrm{E}=\left(\mathrm{E}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{E}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2E}_{\mathrm{1}} \mathrm{E}_{\mathrm{2}} \mathrm{cos2}\theta\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\mathrm{full}\:\mathrm{explanation} \\ $$ Terms of Service Privacy Policy…
Question Number 46374 by Umar last updated on 24/Oct/18 $${At}\:{each}\:{of}\:{three}\:{corners}\:{of}\:{a}\:{square}\: \\ $$$${of}\:{side}\:\mathrm{2}{cm}\:{is}\:{placed}\:{a}\:{point}\:{charge}\: \\ $$$${of}\:{magnitude}\:\mathrm{3}\mu{C}. \\ $$$${What}\:{will}\:{be}\:{the}\:{magnitude}\:{and}\: \\ $$$${direction}\:{of}\:{the}\:{resultant}\:{force}\:{on}\:{a} \\ $$$${point}\:{charge}\:−\mathrm{1}\mu{C}\:{if}\:{it}\:{were}\:{placed} \\ $$$$\:\:\:\left({a}\right)\:\:{at}\:{the}\:{centre}\:{of}\:{the}\:{square}? \\ $$$$\:\:\:\left({b}\right)\:\:{at}\:{the}\:{vacant}\:{corner}\:{of}\:{the}\:{sqre}? \\…