Two point charges Q and -Q/4 are separated by | vedclass

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(D) Let the charges be $q_1 = Q$ at $x=0$ and $q_2 = -Q/4$ at $x=x$.$1$. Potential is zero:$V = \frac{kQ}{r_1} + \frac{k(-Q/4)}{r_2} = 0 \Rightarrow \

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All of the above.

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(D) Let the charges be $q_1 = Q$ at $x=0$ and $q_2 = -Q/4$ at $x=x$.$1$. Potential is zero:$V = \frac{kQ}{r_1} + \frac{k(-Q/4)}{r_2} = 0 \Rightarrow \frac{Q}{r_1} = \frac{Q}{4r_2} \Rightarrow r_1 = 4r_2$.Case $I$: Point between charges. $r_1 + r_2 = x$. $4r_2 + r_2 = x \Rightarrow r_2 = x/5$. This is $x/5$ to the left of $-Q/4$.Case $II$: Point outside charges (closer to smaller magnitude charge). $r_1 - r_2 = x$. $4r_2 - r_2 = x \Rightarrow 3r_2 = x \Rightarrow r_2 = x/3$. This is $x/3$ to the right of $-Q/4$.Both options $A$ and $B$ are correct.$2$. Electric field is zero:$E = \frac{kQ}{r_1^2} - \frac{k(Q/4)}{r_2^2} = 0 \Rightarrow \frac{1}{r_1^2} = \frac{1}{4r_2^2} \Rightarrow r_1 = 2r_2$.For the field to be zero,the point must be outside the charges,closer to the smaller charge. Let the distance from $-Q/4$ be $d$. Then $r_1 = x+d$ and $r_2 = d$.$x+d = 2d \Rightarrow d = x$. This is $x$ to the right of $-Q/4$.Thus,option $C$ is also correct.Since $A, B,$ and $C$ are correct,the answer is $D$.

Two point charges $Q$ and $-Q/4$ are separated by a distance $x$. Then:

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