Question

A block of mass m containing a net negative charge q is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant $\mathrm{k}$ as shown. If horizontal electric field E parallel to the spring is switched on, then the maximum compression of the spring is:

(A) $\sqrt{\dfrac{2 q E}{k}}$
(B) $2 \mathrm{qE} / \mathrm{k}$
(C) $\mathrm{qE} / \mathrm{k}$
(D) Zero

Answer 1

Hint: Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge. Keeping this concept in mind and equating the formula for electric field with spring formula, we can solve the given problem.

Complete step by step solution
We can assume that,
mass of block $=\mathrm{m}$
charge on block $=\mathrm{q}$
spring constant of connected spring to block and wall $=k$
horizontal electric field $=E$
If a horizontal electric field is applied, an accelerating force acts in the right direction on the block. This force is balanced by spring force because spring is also connected to block. Let x be the maximum extension of spring.
So, spring force = electrostatic force
$\Rightarrow \mathrm{kx}=\mathrm{qE}$
$\Rightarrow x=q E / k$
We know, maximum extension of spring is no other than amplitude.
So, amplitude is $x=q E / k$.

Therefore, the correct answer is Option C.

Note: We must keep in mind the formula of electromagnetic field as well as the formula for the work done by a spring in case of solving such questions. We must memorize the values of the constants involved.