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Electricity and Magnetism Level 4

There is a uniformly charged ring having radius $R$ . An infinite line charge (charge per unit length $-\lambda$ , where $\lambda$ is positive constant) is placed along a diameter of the ring (in gravity free space). Total charge on the ring $Q = 4\sqrt { 2 } \lambda R$ .

An electron of mass $m$ and magnitude of charge $e$ is released from rest on the axis of the ring at a distance $x=\sqrt{3} R$ from the centre. Find the initial acceleration of the electron.

\frac { e\lambda }{ \pi { \epsilon }_{ 0 }mR } (\frac { 3+2\sqrt { 2 } }{ 4\sqrt { 3 } } )

None of these choices

\frac { e\lambda }{ \pi { \epsilon }_{ 0 }mR } (\frac { 3+2\sqrt { 2 } }{ 4\sqrt { 6 } } )

\frac { e\lambda }{ \pi { \epsilon }_{ 0 }mR } (\frac { 3-2\sqrt { 2 } }{ 4\sqrt { 6 } } )

1 solution

Rudraksh Shukla
Apr 25, 2016

Electric field due to the wire shall be $E_{wire}=\dfrac{2k\lambda}{\sqrt{3}R}$ And we know that electric field due to a charged ring with charge, radius $Q, R$ at a point with distance $x$ from the center is $E_{ring}=\dfrac{kQx}{(x^2+R^2)^{3/2}}$ Plug in the values given in the question to get $E_{ring}=\dfrac{ek\lambda\sqrt{3}}{R\sqrt{2}}$ We know that $F=qE$ therefore we multiply all the fields with charge $e$ to get the respective forces. Since these are in opposite direction (inward due to ring and outward due to wire) we have $F_{net} =F_{ring} - F_{wire} \Rightarrow F_{net} =\dfrac{e\lambda}{\pi\epsilon{R}} \dfrac{3-2\sqrt{2}}{4\sqrt{6}}$ And by $F=ma$ we have $a=\dfrac{e\lambda}{\pi\epsilon{mR}} \dfrac{3-2\sqrt{2}}{4\sqrt{6}}$

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