two fixed point charges +4e and +e units are separated by a distance a .where should the third point charge be placed for it to be in equilibrium ?


qA = +4e

qB = +e

AB = x

Let, Q be the third point charge located at some point P.

AP = x

therefore, PB = (a-x)

Force on Q due to qA,

F1 = kqAQ/x2

Force on Q due to qB

F2= kQqB/(a-x)2

Since the charge is in equilibrium,

F1 = F2

kqAQ/x2 = k qBQ/(a-x)2

4e/x2 = e/(a-x)2

4/x2 = 1/(a-x)2 (Taking square root on both sides)

2/x = 1/(a-x)

2(a-x) = x(1)

2a - 2x = x

2a = 3x

x = 2/3a

Therefore, the third point charge Q should be placed 2/3a from A.

  • 67
ans can also be x=a/3 
here the third point charge Q should be placed a/3 from B
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Hope this will help you.

  • 63
apply formula and get answer

d (from smaller charge) = a/under root of 4/1 and add 1 get answer
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q should be away by 2/3a from a , the above one , that is perfect answer
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Answer of this question is

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If a is 30 cm then following steps must be followed

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i hope u satishfied

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  • -4
i tink it should be a/3
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sorry I was wrong before, it should be 2a/3 from my calculations
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