Two point charges -20 ,mu C and +40 ,mu C are | vedclass

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(D) The electric potential $V$ due to a point charge $q$ at a distance $x$ is given by $V = \frac{kq}{x}$.For two charges $q_1 = -20 \,\mu C$ and $q_2

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Region $A$ and $B$

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(D) The electric potential $V$ due to a point charge $q$ at a distance $x$ is given by $V = \frac{kq}{x}$.For two charges $q_1 = -20 \,\mu C$ and $q_2 = +40 \,\mu C$ separated by distance $r$,the total potential $V = V_1 + V_2 = 0$.This implies $\frac{kq_1}{x_1} + \frac{kq_2}{x_2} = 0$,or $\frac{|q_1|}{x_1} = \frac{|q_2|}{x_2}$.Since $|q_1| $1$. In Region $A$ (to the left of $-20 \,\mu C$): Let the distance from $-20 \,\mu C$ be $x$. Then $\frac{20}{x} = \frac{40}{r+x} \implies 20r + 20x = 40x \implies 20x = 20r \implies x = r$. Thus,potential is zero in Region $A$.$2$. In Region $B$ (between the charges): Let the distance from $-20 \,\mu C$ be $x$. Then $\frac{20}{x} = \frac{40}{r-x} \implies 20r - 20x = 40x \implies 60x = 20r \implies x = r/3$. Thus,potential is zero in Region $B$.$3$. In Region $C$ (to the right of $+40 \,\mu C$): Since the magnitude of the positive charge is larger,the potential will never be zero here.Therefore,the potential is zero in both Region $A$ and Region $B$.

Two point charges $-20 \,\mu C$ and $+40 \,\mu C$ are separated by a distance $r$. At which region$(s)$ will the electric potential be zero?

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