(Solved): Hi, I don't get the answer from Qb. I tried to draw a vector diagram according to the answer des ...

Hi, I don't get the answer from Qb.

I tried to draw a vector diagram according to the answer description, but I didn't get it. They seem they don't get balanced together...

Can you please help me understand using vector diagrams? (any diagram is fine, I just don't understand only with words)

thank you.

1. In the figure shown, four charges are situated at the corners of a square of side lengers ners are equal to one another and are labeled $$Q$$ for corners $$\mathrm{A}$$ and $$\mathrm{C}$$ and labeled $$q$$ for corners $$\mathrm{B}$$ and $$\mathrm{D}$$. The charge on $$Q$$ is positive for all experiments. (a) In a first experiment, it is found that the force on the charge at position $$\mathrm{C}$$ is $$0\mathrm{N}$$. i. Justify the assertion that the charges $$q$$ cannot have different magnitudes in this experiment. ii. Derive an expression for the magnitude and sign of charge on $$q$$ in this experiment. (b) In a second experiment, the charge at point $$\mathrm{C}$$ is removed. If the charges $$q$$ are positive, is there anywhere within the boundary square where the electric field could be $$0\mathrm{N}/\mathrm{C}$$ ? (c) The charges are reassembled into their original positions. In a clear, coherent paragraph-length response, explain why the electric field at the center of the square must be $$0\mathrm{N}/\mathrm{C}$$ regardless of the magnitudes of the charges $$q$$ or $$Q$$ and regardless of their signs. (b) Within the square, the field from the charge $$Q$$ at position A will point into the fourth quadrant (the $$x$$-component will be positive and the $$y$$-component will be negative). With the charges positive, within the square the field from $$\mathrm{D}$$ will point into the first quadrant (the $$x$$-component will be positive and the $$y$$-component will be positive) and from $$\mathrm{B}$$ will point into the third quadrant (the $$x$$-component will be negative and the $$y$$-component will be negative). To cancel the fields, the $$x$$-coordinate would have to be closer to point $$\mathrm{B}$$ than to point $$\mathrm{D}$$ and the $$y$$-coordinate would have to be closer to $$\mathrm{D}$$ than to B. Some point which satisfies these two requirements will have a field of $$0\mathrm{N}/\mathrm{C}$$.