+4q and -q are two charges separated by a distance d metres . Where should be a third charge +q be placed to attain equilibrium.
Answers
at distance d from -q & 2d from +4q
Explanation:
F = k • Q₁• Q₂ / d²
+4q ------ d ------- -q ---------x ---------- q
Same charges repel , opposite Charges attract
Charges are opposite hence + q to be placed out side .
Let +q Charges must be on right sides of - q .
Let say at distance x
Force on q = k 4q * q / (d+x) ² = k q * q / d²
=> 4d² = (d + x)²
=> d + x = 2d
=> x = d
Hence at distance d from -q & 2d from +4q
or +q Charges on left side of +4q at distance x
Then k 4q * q / (x) ² = k q * q / (d+x)²
=> x² = 4(d + x)²
=> x = 2(d + x)
=> x = 2d + 2x
=> x = - 2d
Again the same situation
at distance d from -q & 2d from +4q
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