Question

Two point charges $+36\mu C$ and $-16\mu C$ are placed 10 cm apart. At a point $X$ , there is no resultant force on unit positive charge. The distance of $X$ from $+36\mu C$ charge is

• A. 10cm
• B. 20cm
• C. 30 cm
• D. 35 cm

Answer 1

Answer: (C)

Both the charges are of opposite signs, so point $X$ will not be in-between the charges.
At point $X,$ there is no resultant force on unit charge, ie, electric field intensity is zero. $E_1=k\frac{36\times {10}^{-6}}{{(10+x)}^2}$ (towards $AX$ ) $E_2=k\frac{16\times {10}^{-6}}{x^2}$ (towards $XB$ or $XA$ ) where, $k=\frac{1}{4\pi {\epsilon }_0}N{m}^2/C^2$ For intensity to be zero at point $X$ . $\therefore$ $E_1=E_2$ $k=\frac{36\times {10}^{-6}}{{(10+x)}^2}=k\frac{16\times {10}^{-6}}{x^2}$ Or $\frac{6}{10+x}=\frac{4}{x}$ $\Rightarrow$ $6x=40+4x$ $x=20cm$ $\therefore$ Distance of point $X$ from $+36\mu C=10+x$ $=10+20=30cm$