Question Number 55431 by necx1 last updated on 24/Feb/19 $$\mathrm{2}\:{plane}\:{parallel}\:{conducting}\:{plates}\:{are} \\ $$$${held}\:{horizontal},{one}\:{above}\:{the}\:{other}\:{in} \\ $$$${a}\:{vacuum}.{Electrons}\:{having}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{6}×\mathrm{10}^{\mathrm{6}} {m}/{s}\:{and}\:{moves}\:{normally}\:{to}\:{the} \\ $$$${plates}\:{enter}\:{the}\:{region}\:{between}\:{them} \\ $$$${through}\:{a}\:{hole}\:{in}\:{the}\:{lower}\:{plate}\:{which} \\ $$$${is}\:{earthed}.{What}\:{potential}\:{must}\:{be} \\ $$$${applied}\:{to}\:{the}\:{other}\:{plate}\:{so}\:{that}\:{the}…
Question Number 52222 by Umar last updated on 04/Jan/19 Answered by peter frank last updated on 04/Jan/19 $$\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…