Question

A block of mass $4\text{kg}$ carrying a charge $50\mu \text{C}$ is connected to a spring for which spring constant is $100\text{N/m}$. The block lies on a frictionless horizontal track and the system is immersed in a uniform electric field of magnitude $5\times {10}^5\text{V/m}$, directed as shown in figure. If the block is released suddenly from rest when the spring is unstretched then the maximum extension in the spring is

• A. $20\text{cm}$
• B. $30\text{cm}$
• C. $40\text{cm}$
• D. $50\text{cm}$

Answer 1

The correct option is D $50\text{cm}$

Given:
$m=4\text{kg};q=50\mu \text{C}=50\times {10}^{-6}\text{C}$

$K=100\text{N/m};E=5\times {10}^5\text{V/m}$


Due to electric field,
force experienced by charge,
$F=qE=50\times {10}^{-6}\times 5\times {10}^5=25\text{N}$

Since, initial block is at rest. So, $v_1=0$
Now again block will come in rest at maximum extension $x$.
So, $v_2=0$

Now apply work energy theorem between initial and final position,

$W_{\text{all forces}}=\Delta KE$

$\Rightarrow \frac{-1}{2}K{x}^2+qEx=K{E}_2-K{E}_1$

$\Rightarrow \frac{-1}{2}K{x}^2+qEx=0(\because v_1=v_2=0)$

$\Rightarrow x=\frac{2qE}{K}=\frac{2\times 25}{100}=0.5\text{m}=50\text{cm}$

Hence, option $\left(d\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - Electrostatic force}(F=qE)\text{is analogue to}\\ \text{Gravitational force}\left(mg\right).\text{Both forces are constant}\\ \text{for a given charge and body. These forces only change the}\\ \text{position of mean position and have no effect on time period of SHM}\end{array}$