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 $

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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



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