Let E-1-r- E-2-r- and E-3-r- be the respective electric fields a distance r from a point charge Q- an infinitely long wire with constant linear charge density lambda- and an infinite plane with uniform surface charge density sigma- If E-1-r-0- - E-2 -r-0- - E-3 -r-0- a given distance r-0- then -Q - 4 sigma pi r-0-2r-0 -frac-lambda-2 pi sigma-E-1 -r-0 - 2- - 2E-2 -r-0-2-E-2 -r-0-2- - 4E-3 -r-0-2-

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Toppr Admin · Accepted Answer

Q4-x3C0-x3B5-0r20-x3BB-2-x3C0-x3B5-0r0-x3C3-2-x3B5-0E1-r02-Q-x3C0-x3B5-0r20-E2-r02-E3-r02-x3C3-2-x3B5-0-x2234-E1-r02-2E2-r02-

List - I	List - II
(P) E is independent of d	(1) A point charge Q at the origin
(Q) $E \propto \frac{1}{d}$	(2) A small dipole with point charges Q at $(0, 0, l)$ and - Q at $(0, 0, - l)$ . Take $2 l << d$
(R) $E \propto \frac{1}{d^{2}}$	(3) A infinite line charge coincident with the x-axis, with uniform linear charge density $λ$ .
(S) $E \propto \frac{1}{d^{3}}$	(4) Two infinite wires carrying uniform linear charge density parallel to the x-axis. The one along ( $y = 0, z = l$ ) has a charge density + $λ$ and the one along $(y = 0, z = - l)$ has a charge density $- λ$ . Take $2 l << d$
	(5) Infinite plane charge coincident with the xy-plane with uniform surface charge density

Q4πε0r20=λ2πε0r0=σ2ε0E1(r02)=Qπε0r20,E2(r02),E3(r02)=σ2ε0∴E1(r02)=2E2(r02)

$\frac{Q}{4 π ε_{0} r_{0}^{2}} = \frac{λ}{2 π ε_{0} r_{0}} = \frac{σ}{2 ε_{0}}$
$E_{1} (\frac{r_{0}}{2}) = \frac{Q}{π ε_{0} r_{0}^{2}}, E_{2} (\frac{r_{0}}{2}), E_{3} (\frac{r_{0}}{2}) = \frac{σ}{2 ε_{0}}$
$∴ E_{1} (\frac{r_{0}}{2}) = 2 E_{2} (\frac{r_{0}}{2})$