question_answer

    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

Correct Answer: C

Solution :

                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, \[k=\frac{1}{4\pi {{\varepsilon }_{0}}}N{{m}^{2}}/{{C}^{2}}\] 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           \[\frac{6}{10+x}=\frac{4}{x}\] \[\Rightarrow \]               \[6x=40+4x\] \[x=20\text{ }cm\] \[\therefore \] Distance of point\[X\]from \[+36\mu C=10+x\] \[=10+20=30\text{ }cm\]