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

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

Electric Charges and Fields

Solution:

$\frac{1}{4 \, \pi \, \varepsilon_0} \frac{2 \times 10^{-6}}{x^2} = \frac{1}{4 \, \pi \, \varepsilon_0} \frac{8 \times b10^6}{(15 - x)^2}$
On solving the obtained Quadratic equation, $x^2 + 10 \,x - 75 = 0$, we get x = 5 cm.