Question

soove it fast ............ ​

Answer 1

${ \bold{ \red{ \underline{ \huge{ ANSWER } : - } }}}$

$\\ \\ { \bold{lets \: \: assume \: \: a \: \: point \: \: where \: \: electric \: \: field}} \\ { \bold{is \: \: zero \: \: have \: \: x \: \: distance \: \: from \: \: 40 \times {10}^{ - 6} c \: .}} \\ \\ { \bold{ \red{ \implies{ { \: E_{1} = E_{2}}}} }}\\ \\ { \bold{ \blue{ \implies \: \frac{k \times 40 \times {10}^{ - 6} }{ {x}^{2} } = \frac{k \times 90 \times {10}^{ - 6} }{ {(4 - x)}^{2} } }}} \\ \\ { \bold{ \blue{ \implies \: \frac{ {(4 - x)}^{2} }{ {x}^{2} } = \frac{9}{4} }}} \\ \\ { \bold{ \red{ \: \: \: \: . \: \: lets \: \: take \: \: square \: \: root \: \: o n\: \: both \: \: side \: - }}} \\ \\ { \bold{ \blue{ \implies \: \frac{4 - x}{x} = \frac{3}{2} }}} \\ \\ { \bold{ \blue{ \implies \:8 - 2x = 3x }}} \\ \\ { \bold{ \blue{ \implies \: 5x = 8}}} \\ \\ { \bold{ \blue{ \implies \:{ \boxed{x = \frac{8}{5}m \: ( \: from \: \: 40 \: micro \: coulomb) }} }}}$

Answer 2

Answer:

$letsassumeapointwhereelectricfieldiszerohavexdistancefrom40×10−6c.⟹E1=E2⟹x2k×40×10−6=(4−x)2k×90×10−6⟹x2(4−x)2=49.letstakesquarerootonbothside−⟹x4−x=23⟹8−2x=3x⟹5x=8⟹x=58m(from40microcoulomb)</p><p>$

Step-by-step explanation:

okay thanks