logo

Ask Questions, Get Answers

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

A substance of dipole moment $4 \times 10^{−8} C\; m$ is placed in an electric field of intensity E. When the direction of the field is changed by an angle of $30^{\circ}$, the heat released by the substance is $2.2\; J.$ What is the value of the electric field intensity E?

Can you answer this question?
 
 

1 Answer

0 votes
$(B) 1.88 \times 10^8 Vm^{-1}$
Hence B is the correct answer.
answered Jun 13, 2014 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...