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प्रश्न
Calculate potential energy of a point charge – q placed along the axis due to a charge +Q uniformly distributed along a ring of radius R. Sketch P.E. as a function of axial distance z from the centre of the ring. Looking at graph, can you see what would happen if – q is displaced slightly from the centre of the ring (along the axis)?
उत्तर
The potential energy (U) of a point charge q placed at-potential V, U = qV In our case a negative charged particle is placed at the axis of a ring having charge Q. Let the ring has radius a, the electric potential at an axial distance z from the centre of the ring is
`V = 1/(4piε_0) Q/sqrt(z^2 + a^2)`
Hence potential energy of a point charge – q is
`U = qV = (-q)[1/(4piε_0) Q/sqrt(z^2 + a^2)]`
⇒ `U = - 1/(4piε_0) (Qq)/sqrt(z^2 + a^2) = 1/(4piε_0) (-Qq)/sqrt(1 + (z/a)^2`
At z = 0, U = `- 1/(4piε_0) (Qq)/a`
At z → `oo`, U → 0
The variation of potential energy with z is shown in the figure. The charge – q displaced would perform oscillations. Nothing can be concluded just by looking at the graph.
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