Question

Two point charges + 9e and te are
separated by a distance 8 m. Where
should another point charge "q" be
placed on the line joining these charges
so that it remains balanced ?​

Answer 1

Explanation:

Now, at equilibrium

Force act on q due to +9e = Force act on q due to +1e

K(+9e)q/r² = k(+1e)q/(16 - r)²

9/r² = 1/(16 - r)²

Taking square root both sides,

3/r = 1/(16 - r)

48 - 3r = r

48 = 4r

r = 12cm

Hence, a point charge q is kept 12cm from +9e

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Answer 2

Answer:

2m

Explanation:

$Force \ on \ charge \ q \ are \ positive \ so \ equarium \ are\ as\\\\Let \ x \ be \ distanceF_{1} =F_{2} \\\\\frac{k(2e)q}{x^{2} } =\frac{k(9e)q}{(8-x)^{2}} \\\\64+x^{2} -16x=9x^{2} \\\\64+8x^{2} -16x=0\\\\x^{2} +2x-8=0\\\\x(x+4)-2(x+4)=0\\\\(x+4)(x-2)=0\\\\x=-4 \ OR \ x=2\ x=-4 \ cannot \ be\\\\So \ your \ answer \ is \ 2$