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
Text Solution
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From following figure,
E A = Electric field at mid point M due to + 5 μ μ C charge
$= 9 \times 10^{9} \times \frac{5 \times 10^{-6}}{(0.1)^2} = 45 \times 10^{5} \, \mathrm{N/C}$
E B = Electric field at M due to +10 μ μ C charge
$= 9 \times 10^{9} \times \frac{10 \times 10^{-6}}{(0.1)^{2}} = 90 \times 10^{5} \, \mathrm{N/C}$
Net electric field at $M = \bigl|\,|\vec{E}_B| - |\vec{E}_A|\,\bigr|$ $= 45 \times 10^{5} \, \mathrm{N/C} = 4.5 \times 10^{6} \, \mathrm{N/C},$
in the direction of E B i.e. towards + 5 μ μ C charge
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