Electrostatics Question 414
Question: Two charges $ +5\mu C $ and $ +10\mu C $ are placed 20 cm apart. The net electric field at the mid-Point between the two charges is [KCET (Med.) 2000]
Options:
A) $ 4.5\times 10^{6} $ N/C directed towards $ +5\mu C $
B) $ 4.5\times 10^{6} $ N/C directed towards $ +10\mu C $
C) $ 13.5\times 10^{6} $ N/C directed towards $ +5\mu C $
D) $ 13.5\times 10^{6} $ N/C directed towards $ +10\mu C $
Show Answer
Answer:
Correct Answer: A
Solution:
$ E _{A} $ = Electric field at mid-point M due to + 5mC charge
$ =9\times 10^{9}\times \frac{5\times {{10}^{-6}}}{{{(0.1)}^{2}}}=45\times 10^{5}N/C $
$ E _{B} $ = Electric field at M due to +10mC charge
$ =9\times 10^{9}\times \frac{10\times {{10}^{-6}}}{{{(0.1)}^{2}}}=90\times 10^{5}N/C $
Net electric field at $ M=|{{\overrightarrow{E}} _{B}}|-|{{\overrightarrow{E}} _{ _{A}}}| $
$ =45\times 10^{5}N/C=4.5\times 10^{6}N/C, $
in the direction of EB i.e. towards + 5mC charge
