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 \[+36\mu C=10+x\] \[=10+20=30\text{ }cm\]You need to login to perform this action.
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