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Question Number 35895 by ajfour last updated on 25/May/18

Commented by ajfour last updated on 25/May/18

$F i n d t h e e l e c t r i c f i e l d d u e t o a$ $u n i f o r m l y c h a r g e d e q u i l a t e r a l$ $t r i a n g u l a r p l a t e (o f c h a r g e$ $d e n s i t y σ) o n t h e y a x i s a t a$ $d i s t a n c e y f r o m o r i g i n (t h e$ $c e n t r o i d o f p l a t e) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/May/18

	$t h e e l e m e n t a r y a r e a a s p e r f i g u r e i s d x d z$ $f o r s i g n c o n v e n i e n c e l e t c o s i d e r t h e p o s i t i o n$ $c o o r i d i n a t e i s (x, z)$ $e l e c t r i c f i e l d a t a d i s t a n c e y a s p e r f i g u r e i s$ $d E = \frac{1}{4 Π ϵ_{0}} \times \frac{σ d x d y}{x^{2} + y^{2} + z^{2}}$ $n o w t o f i n d l i m i t o f x a n d z$ $c e n t r o i d d e v i d e h e i g h t o f t h e t r i a n g l e i n 2 : 1$ $s o d e v i d i n g \frac{\sqrt{3} a}{2} i n 2 : 1 r a t i o w e g e t t h e l i m i t$ $o f z u p p e r l i m i t = \frac{2}{3} \times \frac{\sqrt{3} a}{2}, l o w e r = \frac{1}{3} \times \frac{\sqrt{3} a}{2}$ $Z_{u p} = \frac{a}{\sqrt{3}} Z_{l o w} = \frac{a}{2 \sqrt{3}} l i m i t o f z f r o m \frac{a}{\sqrt{3}} t o (- \frac{a}{2 \sqrt{3}}$ $n o w t o f i n d l i m i t o f x \frac{2}{3} = \frac{x}{\frac{a}{2}}$ $x = \frac{a}{3} l i m i t o f x i s f r o m - \frac{a_{}}{3} t o + \frac{a}{3}$ $E = \frac{1}{4 Π ϵ_{0}} \times \int_{- \frac{a}{3}}^{\frac{a}{3}} \int_{- \frac{a}{2 \sqrt{3}}}^{+ \frac{a}{\sqrt{3}}} \frac{σ d x d z}{x^{2} + y^{2} + z^{2}}$ $g i v e m e t i m e . . . .$
	Commented by ajfour last updated on 25/May/18

	$n o t o k a y, y o u a r e i n e r r o r . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/May/18

	$t r u e$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/May/18

	$w h y y o u h a v e t a k e n d E c o s θ i s t h e r e a n y$ $s y m e t r y t o c a n c e l l t h e d E s i n θ c o m p o n e n t . . .$
	Commented by ajfour last updated on 25/May/18

	$t h e r e i s s u c h a s y m m e t r y .$
	Answered by ajfour last updated on 25/May/18

	$d E = \frac{1}{4 π ϵ_{0}} (\frac{σ d x d z}{x^{2} + y^{2} + z^{2}})$ $E ∣_{(0, y, 0)} = \int \int d E \cos θ$ $= \frac{1}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} [\int_{- \frac{a}{2 \sqrt{3}}}^{(\frac{2}{3} \times \frac{a}{2} - x) \sqrt{3}} \frac{σ y d z}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}}] d x$ $l e t z = \sqrt{x^{2} + y^{2}} \tan ϕ$ $d z = \sqrt{x^{2} + y^{2}} \sec^{2} ϕ d ϕ$ $E = \frac{σ y}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} [\int \frac{\sqrt{x^{2} + y^{2}}}{{(x^{2} + y^{2})}^{3 / 2} \sec^{3} ϕ} \sec^{2} ϕ d ϕ] d x$ $E = \frac{σ y}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} [\int \frac{\cos ϕ d ϕ}{(x^{2} + y^{2})}] d x$ $= \frac{σ y}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} \frac{1}{(x^{2} + y^{2})} {[\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}]}_{- \frac{a}{2 \sqrt{3}}}^{(\frac{a}{3} - x) \sqrt{3}} d x$ $= \frac{σ y}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} \frac{1}{(x^{2} + y^{2})} [\frac{(\frac{a}{3} - x) \sqrt{3}}{\sqrt{x^{2} + y^{2} + 3 {(\frac{a}{3} - x)}^{2}}} + \frac{(\frac{a}{2 \sqrt{3}})}{\sqrt{x^{2} + y^{2} + \frac{a^{2}}{12}}}] d x$ $= \frac{σ y}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} [\frac{(\frac{a}{3} - x) \sqrt{3}}{(x^{2} + y^{2}) \sqrt{4 x^{2} - 2 a x + \frac{a^{2}}{3} + y^{2}}} + \frac{a}{2 \sqrt{3}} \times \frac{1}{(x^{2} + y^{2}) \sqrt{x^{2} + y^{2} + \frac{a^{2}}{12}}}] d x$ $= \frac{σ y}{4 π ϵ_{0}} \int_{- \frac{a}{2}}^{\frac{a}{2}} [\frac{(\frac{a}{3} - x) \sqrt{3}}{(x^{2} + y^{2}) \sqrt{{(2 x - \frac{a}{2})}^{2} + \frac{a^{2}}{12} + y^{2}}} + \frac{a}{2 \sqrt{3}} \times \frac{1}{(x^{2} + y^{2}) \sqrt{x^{2} + y^{2} + \frac{a^{2}}{12}}}] d x$ $. . . . .$
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