Question

Two charges 1 µC and 9 µC are placed at separation of 1 m. Where should a third charge be placed such that, it experience no net force due to these charges?​

Answer 1

Given : Two charges 1 µC and 9 µC are placed at separation of 1 m.

To find : Where should a third charge be placed such that, it experience no net force due to these charges

Solution:

Two charges 1 µC and 9 µC are placed at separation of 1 m.

Third charge should be placed in between both charges so that  it experience no net force due to these charges

as in that case force with be in opposite direction but equal magnitude

Let say charge q distance from 1 µC is   x m

 then distance from 9 µC is  1- x  m

F = kq₁q₂/r²

k(1)q/x²  = k(9)q/(1-x)²

=>  (1 - x)²  = 9x²

=> x² - 2x + 1 =  9x²

=> 8x² + 2x - 1  =0

=> 8x² + 4x - 2x - 1 = 0

=> 4x(2x + 1) - 1(2x + 1) = 0

=> (2x + 1)(4x - 1) = 0

=> x = -1/2  x = 1/4

-ve value not possible as then forces will be in same direction and will not cancel each otther

x = 1/4

1/4 m from charge 1 µC   & 3/4 m from charge 1 µC  

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Answer 2

Answer:

Two charges 1 µC and 9 µC are placed at a separation of 1 m, a third charge is placed such that, it experiences no net force due to these charges, the third charge should place at a distance of $0.25m$ from the first charge

Explanation:

• The electric force due to two point charges ($q_{1} ,q_{2}$) separated by a distance (r) is given by the formula

       $E=K\frac{q_{1} q_{2} }{r^{2} }$

       where the constant $K=9*10^{9}$

From the question, we have two charges arranged as below

Let the third charge be (Q) placed at a distance (x) from the first charge

Force on Q due to $q_{1}$ is $F_{1} =\frac{Kq_{1} Q}{x^{2} }$

Force on Q due to $q_{2}$ is $F_{2} =\frac{Kq_{2} Q}{(1-x)^{2} }$

no net force on the third charge means, the charge is at equilibrium

that is the force on (Q) due to the two charges are equal and opposite

⇒ $\frac{Kq_{1} Q}{x^{2} } =\frac{Kq_{2}Q }{(1-x)^{2} }$

⇒$\frac{K*1*Q}{x^{2} } =\frac{K*9*Q }{(1-x)^{2} }$

⇒$\frac{1}{9} =\frac{x^{2} }{(1-x)^{2} }$

take square root

⇒$1/3=x/1-x$

cross multiply

⇒

$1-x=3x\\1=3x+x=4x\\x=1/4$

the third charge should place at a distance of $0.25m$ from the first charge