Question

please answer this question its very urgent​

Answer 1

Answer: The correct answer is option B

Explanation:

As the 9 micro C charge is having more intensity, so the 3rd charge must be placed as closer as it can be kept near 1 micro C charge because of which the system will get balanced and the net force will be zero

Answer 2

$\large{\underline{\underline{\red{\sf{\hookrightarrow Given:- }}}}}$

• There are two charges of 1μ. and 9μC .
• They are placed 1m apart.

$\large{\underline{\underline{\red{\sf{\hookrightarrow To\:Find:-}}}}}$

• Where we should place third charge , so that it experiences no net force due to these.

$\large{\underline{\underline{\red{\sf{\hookrightarrow Formula\:Used:- }}}}}$

We will use Columb's Law :-

$\large\pink{\underline{\boxed{\purple{\tt{ \dag Force\:\:=\:\:\dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{q_1\:q_2}{r^2}\bigg)}}}}}$

$\large{\underline{\underline{\red{\sf{ \hookrightarrow Answer:-}}}}}$

$\tt[For\: diagram\:refer\:to\: attachment:]$

Let us place that charge (q) at x m distance from 1μC charge .

So , its distance from 9μC charge will be (x-1)m

Now since the net force on q will be 0 , therefore both charges of 1μC and 9μC will exert same force on it .

Let us take that it exerts a force F on it .

$\underline{\blue{\tt Between \:A\:and\:B}}$

$\tt:\implies F = \:\:\dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{q_1\:q_2}{r^2}\bigg)$

$\tt:\implies F= \:\:\dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{1\mu C\times q}{x^2}\bigg)$

$\underline{\boxed{\red{\tt\longmapsto F= \dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{q\mu C }{x^2}\bigg)}}}$

________________________________________

$\underline{\blue{\tt Between \:C\:and\:B}}$

$\tt:\implies F = \:\:\dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{q_1\:q_2}{r^2}\bigg)$

$\tt:\implies F= \:\:\dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{9\mu C\times q}{(1-x)^2}\bigg)$

$\underline{\boxed{\red{\tt\longmapsto F= \dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{9q\mu C }{(1-x)^2}\bigg)}}}$

__________________________________________

$\pink{\tt \leadsto Now \:divide\:these\:two\:equ^n:}$

$\tt:\implies \dfrac{\cancel{F}}{\cancel{F}}=\dfrac{ \dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{9q\mu C }{(1-x)^2}\bigg)}{ \dfrac{1}{4\pi\epsilon_0}\bigg(\dfrac{q\mu C }{x^2}\bigg)}$

$\tt:\implies \cancel{ \dfrac{1}{4\pi\epsilon_0}}\bigg(\dfrac{9q\mu C }{(1-x)^2}\bigg)=\cancel{ \dfrac{1}{4\pi\epsilon_0}}\bigg(\dfrac{q\mu C }{x^2}\bigg)$

$\tt:\implies \dfrac{q}{x^2}=\dfrac{9q}{(1-x)^2}$

$\tt:\implies 9\times x^2=(1-x)^2$

$\tt:\implies 9x^2=1+x^2-2x$

$\tt:\implies 9x^2-x^2+2x+1=0$

$\tt:\implies 8x^2+2x-1=0$

$\tt:\implies 8x^2+4x-2x-1=0$

$\tt:\implies 4x(2x+1)-1(2x+1)=0$

$\tt:\implies (4x-1)(2x+1)=0$

$\underline{\boxed{\red{\tt{\longmapsto \:\:x\:\:=\:\:\dfrac{1}{4},\dfrac{-1}{2}}}}}$

$\bf Since\:x\: can't\;be\; negative\:,so\:x\:=\:\dfrac{1}{4}m$

$\blue{\boxed{\bf{\dag\:\:Hence\:it\:should\:be\:placed\:\dfrac{1}{4}m\:from\:1\mu C\:charge.}}}$