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[1] A thin glass rod is bent into a semicircle of radius $R$. A charge $+Q$ is

uniformly distributed along the upper half and a charge $-Q$ is distributed
uniformly along the lower half as shown in the figure. Find the electric field
at
P, the center of the semicircle.

[2] The electric field due to a line segment of length \(2L\), and carrying a uniform line charge \(\lambda\) at a distance \(d\) above the mid point is given by
\[E = \frac{1}{4\pi\epsilon_0} \, \frac{2\lambda L}{d \sqrt{d^2 +L^2}}\]
Using the above result find the electric field at a distance \(z\) above the
center of a square loop ( side \(2L\)) carrying a uniform line charge
\(\lambda\).

[3] The electric field due to a line segment of length $2a$, and carrying a uniform line charge $\lambda$ at a distance $d$ above the mid point is given by

$$ E = \frac{1}{4\pi\epsilon_0} \, \frac{2\lambda a}{d \sqrt{d^2 +a^2}}$$
Use this result to find the electric field of a {\bf square lamina}
(side $2s$), carrying uniform surface charge density $\sigma$, at a distance
$z$ above the center of the disk. 

[4] The electric field due to a line segment of length \(2a\), and carrying a uniform line charge \(\lambda\) at a distance \(d\) above the mid point is given by
\[E = \frac{1}{4\pi\epsilon_0} \, \frac{2\lambda a}{d \sqrt{d^2 +a^2}}\]
Using the above result find the electric field at a distance \(z\) above the
center of a square loop ( side \(2L\)) carrying a uniform line charge \(\lambda\).

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4727:Diamond Point

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