Question

9c and 16 c separate by 9m wht is zero field location ​

Answer 1

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

$\green{\tt{\therefore{Location=\frac{27}{7}\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Two \: charge \: particle = 9 \: C\: and \: 16 \: C \\ \\ \tt: \implies Distance \: between \: them(r) = 9 \: m \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Place \: where \: electric \: field \: is \: zero = ?$

• According to given question :

$\tt \circ \: q_{1} =9 \: C \: \: \: \: \: \: \: \: \: q_{2} = 16 \: C \\ \\ \tt \circ \: r_{1} = x \: m \: \: \: \: \: \: \: \: \: \: r_{2} =( 9 - x) \: m\\ \\ \bold{As \: we\: know \: that} \\ \tt: \implies E_{1} = E_{ 2} \\ \\ \tt: \implies \frac{K q_{1} }{ { r_{1} }^{2} } = \frac{K q_{2} }{ r_{2} } \\ \\ \tt: \implies \frac{9}{ {x}^{2} } = \frac{16}{(9 - x)^{2} } \\ \\ \tt: \implies 16 {x}^{2} = 9( {9 - x)}^{2} \\ \\ \text{Taking \: underoot \:both \: side} \\ \tt: \implies \sqrt{16 {x}^{2} } = \sqrt{9( 9 - {x)}^{2} } \\ \\ \tt: \implies 4x = 3(9 - x) \\ \\ \tt: \implies 4x = 27 - 3x \\ \\ \tt: \implies 4x + 3x = 27 \\ \\ \green{\tt: \implies x = \frac{27}{7} \: m} \\ \\ \green{\tt \therefore \frac{27}{7} \: m \: \: from \: charge \: q_{1} \: to \: the \: direction \: of \: q_{2}\: net \: electric \: field \: will \: be \: zero}$