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 \[+10\mu C\]
Correct Answer: A
Solution :
From following figure, \[{{E}_{A}}\] = 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 chargeYou need to login to perform this action.
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