A block of mass m having a charge q is placed on a smooth horizontal table and is connected to a wall through an unstressed spring of spring constant k as shown in figure (29-E1). A horizontal electric field E parallel to the spring is switched on. Find he amplitude of he resulting SHM of the block.

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Solution

F = qE, F - kx [Where x amplitude]
Therefore $q E = - k x o r x = \frac{q E}{k}$

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F = qE, F - kx [Where x amplitude] Therefore qE=−kxorx=qEk

F = qE, F - kx [Where x amplitude]
Therefore $q E = - k x o r x = \frac{q E}{k}$