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Electric Field at midpoint of hypotenuse figure

Given, AB = BC = a and Charges at A, B and C are +q, +2q and +q respectively. Found this problem in a textbook. I did it this way :

Charges at A and C will have no effect at this point since electric field due to them at D are equal and acts along opposite directions.

Distance BD will be $$\frac{a}{\sqrt2}$$

Now, electric field due to B on D is given by $$\frac{k.2q}{\left(\frac{a}{\sqrt2}\right)^2}$$ , where k is $$\frac{1}{4\pi\epsilon_0}$$

This gives me the answer : $$\frac{k.4q}{a^2}$$

But the answer given on the textbook is : $$\frac{k.q.2\sqrt2}{a^2}$$

Where am I doing wrong ?

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The real problem here is the slack use of vectors, what you found was the magnitude of electric field but you need to represent it in vector form as the question demands that. Your textbook answer is wrong as well. Change your reference book.
That being said, your calculations are correct.

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  • $\begingroup$ Really , change reference book. I have so far not seen a single perfect book without misprints. $\endgroup$
    – Anubhav Goel
    Commented May 6, 2016 at 12:36
  • $\begingroup$ Misprint would be if the vector representation was incorrect, this mistake showsthat its not a reliable source $\endgroup$
    – Black Jack 21
    Commented May 6, 2016 at 12:38

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