# CBSE Class 12-science Answered

**I don't know which concept and formula is used**

Figure shows that two given charges 4q and -q are separated by a distance 0.4 m .

If we place a third charge and try to get null force on it , the location of third charge right side of negative charge .

If we place the third charge left of +4q charge , positive charge is near to third charge than negative charge .

Hence we always get more repulsive force and less attractive force .

Hence we can not get null force on third charge unless if the third charge is paced at infinity, left side of positive charge .

If we place the third charge in between +4q charge and -q charge , always electorstatic force is directed from third charge to negative charge.

Hence we will never get null force on third charge.

If we place the third charge at right side of -q charge, then we can find a distance x from negative charge

so that attractive force on third charge due to negative charge balances the repulsive force on third charge due to positive charge.

Electrostatic force is proportional to product of charges and inversely prportional to square of distance betwen charges .

Hence we get

4 x^{2} = ( x +0.4)^{2}

3 x^{2} - 0.8 x - 0.16 = 0

Acceptable solution for above quadratic equation is x = 0.4 m

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