Physics help !!?

Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure:

Part A
Specify the strength of the electric field. Take r = 7.1cm .

E= _______ N/C

Part B
Specify the direction.

θ = _________ ∘ below horizontal  

Answer 1

Two point charges with equal and oppsite charge from a equal distance is zero . So now we have to prove it 

Given Two point charges Q1= +3nC

                                         Q2=-3nC

                         

Electric field strength at dot with charge 3 nC {E(+3nC)} is = 

Where K is 8.99 ×10^9 N m2/C2

         d is distance between dot and +3nC point charge r√2 

                                                      ( using pythagoras theorem on right traingles d = )

r = 7.1cm = 0.071m

d = r√2 =  0.071 × 1.414

d = 0.1004m

Now E(+3) = 

= 2675.06

 E(+3) =  2675.06

Ø = arctan (r/r) ⇒ arctan (1) = 45°

E(+3) = r(cosØ + sinØ)

          =2675.06(cos45 + isin45)

          =1891.5 + i1891.5 

Now E(-3) is same as E(+3) but with a different sign 

E(-3) = -1891.5 - i1891.5

Total electric field strength at dot = E(3)+E(-3)

                                                     = 1891.5 + i1891.5 -1891.5 - i1891.5

                                                     = 0

Electric field strength at dot is 0

Comments

There is No direction for the electric field at dot because the electric field strength is '0'

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i think the answer is supose to be 3783.11N/C or 3800N/C and the equation should be

Erod= 1/(4pi * epsilon naut)*/Q/ /r*(r^2*(L/2)^2)^1/2

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whats the angle???