Two charges q and –3q are placed fixed on x-axis separated by distance ‘d’. Where a third charge 2q should be placed such that it will not experience any force?
The charge +2q should be placed to the right of +q charge because if it is placed in between +q and -3q charge both will apply the force in same direction and should be placed close to +q charge because the magnitude is less as compare to +3q charge ,so distance should be less too.
Let suppose it is placed at distance x from +q as shown in fig 1.23 a.

Using Coulomb’s law,
The force on +2q Charge due to +q charge is = k![]()
And the force on +2q Charge due to -3q charge is = k![]()
Where K is Coulombs constant
i.e K= k = 9×109 N �m2 �C−2
So the net force on +2q charge will be
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And according to question net force should be equal to zero.
∴![]()
⇒![]()
⇒
= ![]()
Simplifying it we will get the two value of x
⇒
= ![]()
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Here,
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Now,
x= ![]()
=![]()
=![]()
=![]()
On rationalising we get,
=![]()
=![]()
=![]()
On solving we get
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Now we will exclude the negative value as we have discussed the +2q charge can’t be placed in between.
As d -
will lie in between the charges.So we have to neglect the –ive value,
So ![]()
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