Question

A light weight particle of charge Q is fixed at one end of an electrically insulated uniform elastic rod of natural length L, cross-sectional area A and Young's modulus Y. The rod is placed in space having uniform horizontal electric field of magnitude E and directed parallel to length of the rod as shown. Neglecting gravity and friction then the magnitude of extension in this rod is :

• A. $\frac{QEL}{YA}$
• B. $\frac{QEL}{2YA}$
• C. $\frac{QEL}{4YA}$
• D. None of these

Answer 1

The correct option is A $\frac{QEL}{YA}$

We know that,
$Stress=Y\times strain$
In the question,
$Stress=F/A=QE/A$
$Strain=\delta L/L$
Therefore, $\delta L=QEL/YA$