question_answer

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]

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 charge