Positive charges of $2$ $μ$ C and $8$ $μ$ C are placed $15 c m$ apart. At what distance from the smaller charge will be electric field due to them be zero?

3 c m

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5 c m

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7 c m

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10 c m

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Solution

The correct option is B $5 c m$
Let that distance be $x$ cm
at distance $x$ from the smaller charge, forces from both the charges are equal. Thus,
$\frac{k (2 μ C)}{x^{2}} = \frac{k (8 μ C)}{(15 - x)^{2}}$
$⟹ (15 - x)^{2} = 4 x^{2}$
$⟹ x^{2} + 10 x - 75 = 0$
solving quadratic equation, we get
$x = 5, - 15$

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Positive charges of 2 μC and 8 μC are placed 15cm apart. At what distance from the smaller charge will be electric field due to them be zero?

The correct option is B 5cmLet that distance be x cmat distance x from the smaller charge, forces from both the charges are equal. Thus,k(2μC)x2=k(8μC)(15−x)2⟹(15−x)2=4x2⟹x2+10x−75=0solving quadratic equation, we getx=5,−15

Positive charges of $2$ $μ$ C and $8$ $μ$ C are placed $15 c m$ apart. At what distance from the smaller charge will be electric field due to them be zero?