Answer:b[tex]4.29\times 10^6\ N/C[/tex]
Explanation:
Given
The radius of the ring is [tex]a=7.55\ cm[/tex]
The charge on the ring is [tex]Q=7.06\times 10^{-6}\ C[/tex]
Electric field along the axis at a distance [tex]x[/tex] is given by
[tex]E=\dfrac{kxQ}{(x^2+a^2)^{\frac{3}{2}}}[/tex]
To get the maximum value, differentiate [tex]E[/tex] w.r.t [tex]x[/tex] we get
[tex]E_{max}=\dfrac{2kQ}{3\sqrt{3}a^2}\quad [\text{at}\ x=\dfrac{a}{\sqrt{2}}]\\\\E_{max}=\dfrac{2\times 9\times 10^9\times 7.06\times 10^{-6}}{3\sqrt{3}(7.55\times 10^{-2})^2}\\\\E_{max}=\dfrac{24.456\times 10^3}{57\times 10^{-4}}=4.29\times 10^6\ N/C[/tex]
