Question

A piece of wire has the shape of a semicircle with radius 𝑎 and lies in the first and second quadrant.

The wire is charged with a (line) charge density 𝜆(𝜃)=𝛼sin𝜃 with 𝜃 the angle and 𝛼 a constant.

Calculate the total wire charge 𝑄.

Why is the radius of this semicircle involved? I tried integrating over the angle from 0 to pi, but this gives me the answer 2𝛼, while the answer should be 2a𝛼.

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Answer 1

 𝜆(𝜃)= 𝛼sin𝜃 

$dQ= \alpha sin(\theta) adth\eta$

$Q = \intop^{\pi/2}_{0} a\alpha sin(\theta)$

$\boxed {Q = \alpha a }$

as you solved this is answer

you are saying that answer is $2 \alpha a$

than

it must be solved like this

this is due to

horizontal component 

dEsinθ will be cancelled out with opposite element, and 

dEcosθ components get added up

So

$dQ= \alpha cos(\theta) adth\eta$

$Q = \intop^{\pi/2}_{-\pi /2} a\alpha cos(\theta)$

$Q = [sin(\theta)]^{\pi/2}_{-\pi/2}$

$\boxed {Q = 2\alpha a }$