Two charges +5,mu C and +10,mu C are placed 2 | vedclass

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Question

(A) Let the charges be $q_A = +5\,\mu C$ and $q_B = +10\,\mu C$ placed at points $A$ and $B$ respectively,separated by a distance $d = 20\,cm = 0.2\,m

Vedclass · Accepted Answer

$4.5 	imes 10^6\,N/C$ directed towards $+5\,\mu C$

Vedclass · Answer

(A) Let the charges be $q_A = +5\,\mu C$ and $q_B = +10\,\mu C$ placed at points $A$ and $B$ respectively,separated by a distance $d = 20\,cm = 0.2\,m$.The midpoint $M$ is at a distance $r = 10\,cm = 0.1\,m$ from both charges.The electric field $E_A$ at $M$ due to $q_A$ is directed away from $A$ (towards $B$):$E_A = \frac{k|q_A|}{r^2} = \frac{9 	imes 10^9 	imes 5 	imes 10^{-6}}{(0.1)^2} = 45 	imes 10^5\,N/C = 4.5 	imes 10^6\,N/C$.The electric field $E_B$ at $M$ due to $q_B$ is directed away from $B$ (towards $A$):$E_B = \frac{k|q_B|}{r^2} = \frac{9 	imes 10^9 	imes 10 	imes 10^{-6}}{(0.1)^2} = 90 	imes 10^5\,N/C = 9.0 	imes 10^6\,N/C$.Since $E_A$ and $E_B$ are in opposite directions,the net electric field $E_{net} = |E_B - E_A| = 9.0 	imes 10^6 - 4.5 	imes 10^6 = 4.5 	imes 10^6\,N/C$.Since $E_B > E_A$,the net field is directed in the direction of $E_B$,which is towards the $+5\,\mu C$ charge.

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

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