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Equation of Continuity

Two point cha...

Question

Two point charges $q_{1} = 4 μ C$ and $q_{2} = 9 μ C$ are placed $20 c m$ apart. The electric field due to them will be zero on the line joining them at a distance of

8 c m

from

q_{1}

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8 c m

from

q_{2}

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\frac{80}{13} c m

from

q_{1}

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\frac{80}{13} c m

from

q_{2}

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Solution

The correct option is A $8 c m$ from $q_{1}$
Let's assume electric field is zero at a distance $x$ from first point charge.
Then at the point where electric field will be zero:
$E_{p} = 0 \Rightarrow \frac{4}{x^{2}} = \frac{9}{{(20 - x)}^{2}} \Rightarrow \frac{20 - x}{x} = \frac{3}{2}$
$\Rightarrow 40 - 2 x = 3 x \Rightarrow x = 8 c m$

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