Question Number 36851 by Tinkutara last updated on 06/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18 $$\bigtriangleup{V}={V}_{\mathrm{2}} −{V}_{\mathrm{1}} =\mathrm{80}−\mathrm{120}=−\mathrm{40} \\ $$$$\bigtriangleup{r}={r}_{\mathrm{2}} −{r}_{\mathrm{1}} =\mathrm{1}−\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${E}=−\frac{\bigtriangleup{V}}{\bigtriangleup{r}}=\frac{−\left(−\mathrm{40}\right)}{\mathrm{2}}=\mathrm{20}…
Question Number 36722 by Tinkutara last updated on 04/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18 $${the}\:{eletric}\:{field}\:{is}\:\overset{\rightarrow} {{E}}={E}\hat {{i}} \\ $$$${let}\:{assume}\:\:{a}\:{positive}\:{charge}\:{Q}\:{is}\:{situated} \\ $$$${at}\:{point}\:{p}\left(−{l},\mathrm{0}\right)\:{which}\:{produce}\:{Electric}\:{field} \\ $$$${now}\:{potential}\:{at}\:{point}\:{A}\:{is}\:{V}_{{A}}…
Question Number 36096 by rahul 19 last updated on 28/May/18 Commented by rahul 19 last updated on 30/May/18 $$?? \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 35940 by ajfour last updated on 26/May/18 Commented by ajfour last updated on 26/May/18 $${Find}\:{electric}\:{field}\:{at}\:\boldsymbol{{P}}\:{due}\:{to}\:{a} \\ $$$${a}\:{charged}\:{rectangular}\:{plate}\:{of} \\ $$$${surface}\:{charge}\:{density}\:\boldsymbol{\sigma}. \\ $$ Answered by…
Question Number 35895 by ajfour last updated on 25/May/18 Commented by ajfour last updated on 25/May/18 $${Find}\:{the}\:{electric}\:{field}\:{due}\:{to}\:{a} \\ $$$${uniformly}\:{charged}\:{equilateral} \\ $$$${triangular}\:{plate}\:\left({of}\:{charge}\:\right. \\ $$$$\left.{density}\:\sigma\right)\:{on}\:{the}\:{y}\:{axis}\:{at}\:{a}\: \\ $$$${distance}\:{y}\:{from}\:{origin}\:\left({the}\right.…
Question Number 35851 by ajfour last updated on 24/May/18 Commented by ajfour last updated on 25/May/18 $${Q}.\mathrm{35828}\:{continuation}.. \\ $$$${Find}\:{electric}\:{potential}\:{at}\:{point}\: \\ $$$${P}\left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} ,{z}_{\mathrm{0}} \right)\:{due}\:{to}\:{the}\:{charged} \\…
Question Number 35828 by ajfour last updated on 24/May/18 Commented by ajfour last updated on 24/May/18 $${Find}\:{the}\:{electric}\:{field}\:{at}\:{point}\:{P} \\ $$$${on}\:{the}\:{y}\:{axis}\:{at}\:{a}\:{distance}\:{y}\:{from} \\ $$$${origin}\:{due}\:{to}\:{a}\:{uniformly}\:{charged} \\ $$$${square}\:{wire}\:{frame}\:{of}\:{linear}\:{charge} \\ $$$${density}\:\lambda.\:{The}\:{frame}\:{lies}\:{in}\:{xz}…
Question Number 34652 by 33 last updated on 09/May/18 $${a}\:{point}\:{charge}\:{q}\:{is}\:{placed} \\ $$$${at}\:\left(\mathrm{0},{a}/\mathrm{2},\mathrm{0}\right)\:.\:{find}\:{flux} \\ $$$${due}\:{to}\:{point}\:{charge}\:{through} \\ $$$${a}\:{square}\:{sheet}\:{of}\:{side}\:{a}\:\: \\ $$$${in}\:{xz}\:{plane}\:{whose}\:{centre} \\ $$$${coincides}\:{with}\:{orign}. \\ $$$$\left(\:{do}\:{not}\:{use}\:{gauss}\:{law}\right) \\ $$ Commented…
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…