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Question

Two charges are placed. Let, the third charge 9 e placed at a distance $	imes$ from 16 e and hence its distance from $9 e$ is $(70- x ) cm$ for the s

Vedclass · Accepted Answer

$(a)$ અને $(b)$ બંને

Vedclass · Answer

Two charges are placed. Let, the third charge 9 e placed at a distance $	imes$ from 16 e and hence its distance from $9 e$ is $(70- x ) cm$ for the system to remain in equilibrium $F _1= F _2$

$	ext { or, } F_1=K 	imes \frac{9 e 	imes q}{x^2}$

$F_2=\frac{K 	imes 16 e 	imes q}{(70-x)^2}$

Equating $F _1$ and $F _2$ and we get,

Now,

$F_1=F_2$

$	ext { or, } \frac{9 e 	imes q}{x^2}=\frac{16 e 	imes q}{(70-x)^2}$

$	ext { or, } \frac{3}{x}=\frac{4}{(70-x)^2}$

or, $3(70-x)=4 x$

or, $7 x=48$

or, $210-3 x =4 x$

or, $x =\frac{210}{7}=30\,cm$

now $30\,cm$ from $9\,e$.

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