+4q and -q are two charges separated by a distance "d' metre. Where should be a third charge + Q be placed to attain equilibrium?
Answers
Given :
Two charges +4q and -q.
The distance between the two charges = d metres.
To find :
Location of the third charge so that it attains equilibrium.
Solution :
Let the charge +Q is placed at distance x from charge -q.
The distance of +Q from +4q = x + d
For equilibrium, the net force on the charge +Q should be zero.
[ k * (+Q) * (+4q) / (x+d) * (x+d) ] + [ k * (+Q) * (-q) / (x) * (x) ] = 0
=> 4 =
=> 2 = (x +d) / x
=> 2x = x + d
=> x = d metres
The third charge is placed at d metres from charge -q and at distance 2d from the charge +4q .
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
-2d on left side from 4q means 2d right side of 4q
hence d right of -q
Again the same situation
at distance d from -q & 2d from +4q
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