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

• 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 $+ 10\mu C$

Answer 1

Answer: (A)

$4.5 \times 10^{6}$ N/C directed towards $+ 5\mu C$

Answer 2

From following figure,

E_A = Electric field at mid point M due to + 5μC 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 +10μC charge

$= 9 \times 10^{9} \times \frac{10 \times 10^{- 6}}{(0.1)^{2}} = 90 \times 10^{5}N ⥂ / ⥂ C$

Net electric field at$M = \left| \vec { E } _ { B } \right| - \left| \vec { E } _ { A } \right|$

$= 45 \times 10^{5}N ⥂ / ⥂ C = 4.5 \times 10^{6}N ⥂ / ⥂ C,$

in the direction of E_B i.e. towards + 5μC charge