Physics, asked by Gowrilechu, 10 months ago

Two points charges of 204 C and
80 k c are
placed 18 cm apart.
Find the position where the electric
Peeld is hero?​

Answers

Answered by BrainlyConqueror0901
10

CORRECT QUESTION :

Two points charges of 20 u C and 80 u c are placed 18 cm apart. Find the position where the electric field is zero?

\blue{\bold{\underline{\underline{Answer:}}}}

\green{\tt{\therefore{Place\:where\:electric\:field\:is\:zero=6\:cm\:(From\:20\:\mu\:C\:charge)}}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

 \green{\underline \bold{Given :}} \\  \tt: \implies Two \: points \: charges = 20 \mu C\\\\ \tt:\implies Distance\:between\:charges=18\:cm\\\\ \red{\underline \bold{To\:Find :}}\\\tt:\implies Place\:where\:electric\:field\:is\:zero=?

• According to given question :

 \tt \circ \:  q_{1}  = 20  \:  \mu \: C\:  \:  \:  \:  \:  \:  q_{2} = 80 \:  \mu \: C \\  \\  \tt \circ \:  r_{1}  = (0.18 - x) \: m \:  \:  \:  \:  \:  \:  r_{2}  =x  \: m\\\\\ \bold{At \:Point\: A:} \\  \tt: \implies  F_{1} =  F_{2} \\  \\ \tt: \implies  \frac{k q_{1} }{ ({ r_{1}})^{2} }  =  \frac{k q_{2} }{ {( r_{2} )}^{2} }  \\  \\ \tt: \implies  \frac{20 \times  {10}^{ - 6} }{ {(0.18 - x)}^{2} }  =  \frac{80 \times  {10}^{ - 6} }{ {(x)}^{2} }  \\  \\ \tt: \implies{x}^{2}  =  \frac{80}{20}  \times  {(0.18 - x)}^{2}  \\  \\ \tt: \implies x^{2}  = 4 \times  ({0.18 - x)}^{2}  \\  \\  \text{Taking \: under\:root \: both\:side:} \\   \tt:\implies \sqrt{(x)^{2} }  =  \sqrt{4 \times  {(0.18 - x)}^{2} }  \\  \\ \tt: \implies x = 2 \times (0.18 - x) \\  \\ \tt: \implies x = 0.36 - 2x \\  \\ \tt: \implies 3x = 0.36 \\  \\  \green{\tt: \implies x = 0.12 \: m=12cm\:(From\:80\:\mu\:C\:charge)} \\  \\   \green{\tt \therefore Place \: where \: electric \: field \: is \: zero \: is \:6 \: cm \: from \: 20  \:  \mu \: c \: charge}

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Answered by Anonymous
9

Explanation:

Correct Question :

Two points charges of 20 u C and 80 u c are placed 18 cm apart. Find the position where the electric field is zero?

ANSWER :

Position where electric field is zero = 9cm

GIVEN :

==> Two Point charges 20μC and 80μC

==> Distance between charges = 18 cm

According to Question

q1 = 20 μC ; q2 =80μC

r1 = (0.18 - x) cm ; r2 = 0.18 cm

At Point A

==> F1 = F2

==> \:  \:  \frac{kq1}{ {r1}^{2} }   = \frac{kq2}{ {r2}^{2}   }  \\  \\ ==> \:  \frac{20 \times  {10}^{ - 6} }{(0.18 - x)}  \:  \:  =  \:  \:  \frac{80 \times  {10}^{ - 6} }{0.18}  \\  \\ ==> \:  {0.18}^{2}  =  \frac{80}{20}  {(0.18 - x)}^{2}  \\  \\ ==> \:  {0.18}^{2}  = 4 {(0.18 - x)}^{2}  \\  \\  \\ taking \: underroot \: both \: sides \: we \: get \:  \\  \\==> \sqrt{ {0.18}^{2} }  =  \sqrt{4 {(0.18 - x)}^{2} }  \\  \\ ==> \: 0.18 = 2(0.18 - x) \\  \\ ==> \: 0.09 = 0.18 - x \:  \\  \\ ==>  x \:  =  \: 0.18 - 0.09 = 0.09 \\  \\ ==> \: x \:  =  \: 0.09 \\  \\ ==> \: x \:  =  \: 9cm

Hence ,

Position where electric field is zero is 9cm from 20μC

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