91Ó°ÊÓ

We know that the electric field of a thin ring, like that on the above figure at point \(P\) is given by:

\(E = \frac{{kQx}}{{{{({x^2} + {a^2})}^{\frac{3}{2}}}}}\)

Where, \(Q\) is the total charge and a is the radius. Now at an infinity distance from the canter of the ring, the value of \({a^2}\) will be negligible. Therefore, the equation can also be written as:

\(E = \frac{{kQx}}{{{{({x^2})}^{\frac{3}{2}}}}}\)

\(\begin{aligned}E = \frac{{kQx}}{{{{({x^2})}^{\frac{3}{2}}}}}\\ = \frac{{kQx}}{{{x^3}}}\end{aligned}\)

And finally, the formula can also be written in such a way that it gives the electric field of a point charge given by:

\(E = \frac{{kq}}{{{r^2}}}\)

Now \(E{x^2} = kQ\), and since k is constant and the total charge of the ring \(Q\) is also constant, hence the value of \(kQ\) is also constant. This is possible only at huge values of \(x\).

Therefore: \(E{x^2} = cons\tan t\)

From figure we get that \(E{x^2}\)is \(45\), therefore

\(45 = kQ\)

Now solving for \(Q\) we get:

\(\begin{aligned}Q = \frac{{45}}{k}\\ = \frac{{45}}{{9.0 \times {{10}^9}}}\\ = 5.0 \times {10^{ - 9}}\end{aligned}\)

Therefore, the charge is \(5.0 \times {10^{ - 9}}\)C

Again, the electric field of the ring is given by:

\(E = \frac{{kQx}}{{{{({x^2} + {a^2})}^{\frac{3}{2}}}}}\)

Now when the \(x\) become too small the value of \({x^2}\)will be negligible. There the equation can also be written as:

\(\begin{aligned}E = \frac{{kQx}}{{{{({a^2})}^{\frac{3}{2}}}}}\\ = \frac{{kQx}}{{{a^3}}}\end{aligned}\)

Now dividing both side by\(x\)we get:

\(\frac{E}{x} = \frac{{kQ}}{{{a^3}}}\)

Since \(k,Q,a\) are constant therefore: \(\frac{E}{x} = cons\tan t\)

Putting the values and solving for a we get:

\(\begin{aligned}700 = \frac{{kQ}}{{{a^3}}}\\{a^3} = \frac{{kQ}}{{700}}\\a = \sqrt(3){{\frac{{kQ}}{{700}}}}\\a = \sqrt({}){{\frac{{9.0 \times {{10}^9} \times 5.0 \times {{10}^{ - 9}}}}{{700}}}}\\ = 0.40m\end{aligned}\)

Therefore, the value of a is\(0.40m\)