Question

Two charges $+5\mu C$ and $+10\mu C$ are placed 20 cm apart. The electric field at the midpoint between the two charges is:

• A. $4.5\times {10}^6$ N/C towards s $+5\mu C$
• B. $4.5\times {10}^6$ N/C towards $+10\mu C$
• C. $13.5\times {10}^6$ N/C towards $+5\mu C$
• D. $13.5\times {10}^6$ N/C towards $+10\mu C$

Answer 1

The correct option is C $13.5\times {10}^6$ N/C towards $+5\mu C$

Let the two charges placed at two points of line AB as A and B whose distance is 20 cm and O is the mid point of the line AB

Distance between the two charges , $AB=20cm$

$\therefore AO=OB$

Net elctric at O$=E$

Electric field at point O caused by +5\mu C charge,

$E_1=\frac{5\times {10}^{-6}}{4\pi {\epsilon }_0(AO{)}^2}$

$=\frac{5\times {10}^{-6}}{4\pi {\epsilon }_0(10\times {10}^{-2}{)}^2}N/Ctoward+5\mu C$

where,

${\epsilon }_0=$Permittivity of free space

$\frac{1}{4\pi \epsilon }=9\times {10}^{9}N{m}^2{C}^{-2}$

Magnitude of electric field at point O cause by +10\mu C charge,

$E_1=\frac{10\times {10}^{-6}}{4\pi {\epsilon }_0(AO{)}^2}$

$=\frac{10\times {10}^{-6}}{4\pi {\epsilon }_0(10\times {10}^{-2}{)}^2}N/Ctowards+10\mu C$

$\therefore E=E_1+E_2$

$=\frac{9\times {10}^9\times 15\times {10}^{-6}}{(10\times {10}^{-2}{)}^2}$

$=13.5\times {10}^{6}N/Ctowards+5\mu C$