A block of mass m with 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 the figure. A horizontal electric field E, parallel to the spring, is switched on. Find the amplitude of the resulting SHM of the block.

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Solution

Given:
Charge of the block = q
Electrical field intensity = E
Mass of the block = m
Spring constant = k
Electric force, F = qE
Spring force, F = − kx (where x is the amplitude)
Therefore, qE = − kx
$\Rightarrow x = |- \frac{q E}{k}| = \frac{q E}{k}$

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A block of mass m with 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 the figure. A horizontal electric field E, parallel to the spring, is switched on. Find the amplitude of the resulting SHM of the block.

Given: Charge of the block = q Electrical field intensity = E Mass of the block = m Spring constant = k Electric force, F = qE Spring force, F = − kx (where x is the amplitude) Therefore, qE = − kx ⇒x=-qEk=qEk

Given:
Charge of the block = q
Electrical field intensity = E
Mass of the block = m
Spring constant = k
Electric force, F = qE
Spring force, F = − kx (where x is the amplitude)
Therefore, qE = − kx
$\Rightarrow x = |- \frac{q E}{k}| = \frac{q E}{k}$