Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A 10.0 Kg block carrying a charge 100 $\;\mu C\;$ is connected to a spring for which $\;K=100 N/m\;$ . The block lies on a frictionless horizontal track , and the system is immersed in a uniform electric field $\;E=5\times10^{5}\;V/m\;$ , directed as shown in figure . If the block is released from rest when the spring is amount does the spring expand

$(a)\;0.5 m \qquad(b)\; 1 m\qquad(c)\;0.25 m\qquad(d)\;2 m$

Can you answer this question?

1 Answer

0 votes
Answer : (b) 1 m
Explanation :
Work done by electric field = Work done against spring
Let $\;W_{E}\;$ be the work done by electric field . Then
Work done by spring = - $\large\frac{1}{2}\;k\;x^2$
Therefore $\;Q\;E\;x-\large\frac{1}{2}\;k\;x^2= \bigtriangleup K.E$
For maximum expansion $\;K.E_{final} =0$
But $\;K.E_{initial}=0$
Then $\;\bigtriangleup K. E=0$
Thus $\;Q\;E\;x-\large\frac{1}{2}\;k\;x^2=0$
answered Feb 14, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App