n identical point charges are kept symmetricall on the periphery of the circle $x^{2}+y^{2}=R^{2}$ in $x y$…
n identical point charges are kept symmetricall on the periphery of the circle $x^{2}+y^{2}=R^{2}$ in $x y$ plane. The resultant electric field at $(0,0, \mathrm{R})$ is $\mathrm{E}_{1}$ and at $(0,0,2 \mathrm{R})$ is $\mathrm{E}_{2}$. The ratio of $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}$ is
$\frac{5 \sqrt{5}}{4 \sqrt{2}}$
$\frac{5}{2}$
$\frac{5}{4}$
$\frac{5 \sqrt{5}}{2 \sqrt{2}}$
Solution
At \((0,0, R)\) :
1. Distance from any charge to \((0,0, R): d=R \sqrt{2}\).
2. Electric field due to one charge: \(E=\frac{k q}{(R \sqrt{2})^2}=\frac{k q}{2 R^2}\).
3. Vertical component: \(E_z=\frac{E}{\sqrt{2}}=\frac{k q}{2 R^2 \sqrt{2}}\).
4. Total electric field: \(E_1=n \cdot \frac{k q}{2 R^2 \sqrt{2}}=\frac{n k q}{2 R^2 \sqrt{2}}\).
At \((0,0,2 R)\) :
1. Distance from any charge to \((0,0,2 R): d=R \sqrt{5}\).
2. Electric field due to one charge: \(E=\frac{k q}{(R \sqrt{5})^2}=\frac{k q}{5 R^2}\).
3. Vertical component: \(E_z=\frac{2 E}{\sqrt{5}}=\frac{2 k q}{5 R^2 \sqrt{5}}\).
4. Total electric field: \(E_2=n \cdot \frac{2 k q}{5 R^2 \sqrt{5}}=\frac{2 n k q}{5 R^2 \sqrt{5}}\).
Ratio \(\frac{E_1}{E_2}\) :
\(\frac{E_1}{E_2}=\frac{\frac{n k q}{2 R^2 \sqrt{2}}}{\frac{2 R^2 q}{5 R^2 \sqrt{5}}}=\frac{5 \sqrt{5}}{4 \sqrt{2}}\)
Therefore, the ratio \(\frac{E_1}{E_2}\) is:
\(\frac{E_1}{E_2}=\frac{5 \sqrt{5}}{4 \sqrt{2}}\)
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