JEE Main & Advanced
JEE Main Paper (Held On 09-Jan-2019 Morning)
question_answer
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:
[JEE Main Online Paper (Held On 09-Jan-2019 Morning]
A)\[R\sqrt{2}\]
B)R
C)\[\frac{R}{\sqrt{5}}\]
D)\[\frac{R}{\sqrt{2}}\]
Correct Answer:
D
Solution :
\[E\,\,\,\,=\,\,\,\,\frac{kQx}{{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\] For \[{{E}_{\max }}\] \[\frac{dE}{dx}\,\,=\,\,0\] On solving we get \[x\,\,=\,\,\frac{R}{\sqrt{2}}\] So, option is correct.