Charges of 2 mu C and -3 mu C are placed at t | vedclass

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(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{Kq}{r}$.Let $r_1$ be the distance from point $A$ (wh

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$0.4$

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(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{Kq}{r}$.Let $r_1$ be the distance from point $A$ (where charge $q_1 = 2 \mu C$) and $r_2$ be the distance from point $B$ (where charge $q_2 = -3 \mu C$) to the point where the net potential is zero.For the net potential to be zero,the sum of potentials must be zero: $V_A + V_B = 0$.$\frac{K(2 	imes 10^{-6})}{r_1} + \frac{K(-3 	imes 10^{-6})}{r_2} = 0$$\frac{2}{r_1} = \frac{3}{r_2}$$\frac{r_2}{r_1} = \frac{3}{2}$Since the total distance between $A$ and $B$ is $1 \ m$,we have $r_1 + r_2 = 1 \ m$,which implies $r_2 = 1 - r_1$.Substituting this into the ratio:$\frac{1 - r_1}{r_1} = \frac{3}{2}$$2(1 - r_1) = 3r_1$$2 - 2r_1 = 3r_1$$5r_1 = 2$$r_1 = \frac{2}{5} = 0.4 \ m$.Thus,the distance from point $A$ is $0.4 \ m$.

Charges of $2 \mu C$ and $-3 \mu C$ are placed at two points $A$ and $B$ separated by $1 \ m$. The distance of the point from $A$,where the net potential is zero,is (in $m$)

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