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Principle of Superposition

Two positive ...

Question

Two positive charges $q_{1} = 4 \times 10^{- 6} C$ and $q_{2} = 9 \times 10^{- 6} C$ are placed $10 c m$ apart in air. The position of a third charge to be placed between them, such that there will be no resultant force on it is:

6 c m

from

q_{1}

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3 c m

from

q_{1}

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4 c m

from

q_{1}

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7 c m

from

q_{1}

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Solution

The correct option is C $4 c m$ from $q_{1}$
Net force on Q must be zero.
So, $F_{n e t} = \frac{k q_{1} (Q)}{x^{2}} - \frac{k q_{2} (Q)}{(10 - x)^{2}} = 0$
$\Rightarrow \frac{q_{1}}{x^{2}} - \frac{q_{2}}{(10 - x)^{2}} = 0$
$\Rightarrow \frac{4 \times 10^{- 6}}{x^{2}} = \frac{9 \times 10^{- 6}}{(10 - x)^{2}}$
$\Rightarrow 2 (10 - x) = 3 x$
$\Rightarrow 20 = 5 x$
$\Rightarrow x = 4 c m$

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Two positive charges q1=4×10−6 C and q2=9×10−6 C are placed 10 cm apart in air. The position of a third charge to be placed between them, such that there will be no resultant force on it is:

The correct option is C 4 cm from q1Net force on Q must be zero.So, Fnet=kq1(Q)x2−kq2(Q)(10−x)2=0⇒q1x2−q2(10−x)2=0⇒4×10−6x2=9×10−6(10−x)2⇒2(10−x)=3x⇒20=5x⇒x=4 cm

Two positive charges $q_{1} = 4 \times 10^{- 6} C$ and $q_{2} = 9 \times 10^{- 6} C$ are placed $10 c m$ apart in air. The position of a third charge to be placed between them, such that there will be no resultant force on it is: