A simple problem

Place an opaque "blob" of charge on a transparent, hemispherical nonconducting shell, as shown. . Find z component of field due to it, at centre O.

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Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Is it $\vec{E} = \dfrac{\sigma A}{4\pi\epsilon_0 r^2}\hat{e_z}$ ?

Kishore S. Shenoy - 4 years, 3 months ago

@Kishore S Shenoy (vector = scalar ?) Yeah, you are right. The result is trivial, but has applications. For example, if we want to calculate field due to a hemispherical shell of charge at the centre, without messy integration.

Venkata Karthik Bandaru - 4 years, 3 months ago

Yeah, I know. That's the beauty of nature's dimensional analogy!

Kishore S. Shenoy - 4 years, 3 months ago

How can I calculate it

A Former Brilliant Member - 2 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$