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Question Number 36722 by Tinkutara last updated on 04/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

$$theeletricfieldis\overrightarrow{E}=E\hat{i}$$$$letassumeapositivechargeQissituated$$$$atpointp(-l,0)whichproduceElectricfield$$$$nowpotentialatpointAis{V}_A$$$$V_A=\frac{Q}{4\mathrm{\Pi }{ϵ}_0(l+a)}=\frac{Q}{4\mathrm{\Pi }{ϵ}_0\left(\sqrt{l^2+a^2+2la}\right)}$$$$V_B=\frac{Q}{4\mathrm{\Pi }{ϵ}_0\left(\sqrt{l^2+a^2}\right)}=\frac{Q}{4\mathrm{\Pi }{ϵ}_0(\sqrt{l^2+a^2)}}$$$$V_c=\frac{Q}{4\mathrm{\Pi }{ϵ}_0(l-a)}=\frac{Q}{4\mathrm{\Pi }{ϵ}_0\sqrt{l^2+a^2-2la}}$$$$V_D=\frac{Q}{4\mathrm{\Pi }{ϵ}_0\left(\sqrt{l^2+a^2}\right)}=\frac{Q}{4\mathrm{\Pi }{ϵ}_0\sqrt{l^2+a^2}}$$$$so{V}_{A}isminimum$$$$V_{c}ismaximum$$$$V_B=V_D$$

Commented by Tinkutara last updated on 06/Jun/18

Thank you very much Sir! I got the answer. ��������

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