Question
(a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.53, given that qa=qb=−1.00 μCq_a = q_b = -1.00 \textrm{ }\mu\textrm{C} and qc=qd=+1.00 μCq_c = q_d = +1.00 \textrm{ }\mu\textrm{C}. (b) Calculate the magnitude of the electric field at the location of qq, given that the square is 5.00 cm on a side.
<b>Figure 18.53</b> Four charges distributed at the corners of a square, with a charge in the center.
Figure 18.53 Four charges distributed at the corners of a square, with a charge in the center.
Question by OpenStax is licensed under CC BY 4.0
Final Answer

2.03×107 N/C, up2.03 \times 10^7 \textrm{ N/C, up}

Solution video

OpenStax College Physics, Chapter 18, Problem 46 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A force of 4.80 times 10 to the minus 17 newtons is exerted to the west on an electron; the charge on an electron is the elementary charge— 1.60 times 10 to the minus 19 coulombs— and our question is what is the electric field? So in order for the force to be to the left that means the field must be to the right because the field direction is in the same direction as the force would be on a positive test charge. So this being negative, it will be a force in the opposite direction. So the force magnitude is the charge multiplied by the electric field and so we divide both sides by q to solve for E. So electric field then is 4.80 times 10 to the minus 17 newtons divided by 1.60 times 10 to the minus 19 coulombs and that's 300 newtons per coulomb and so our final answer is 3.00 times 10 to the 2 newtons per coulomb to the east is the electric field and so that's to the right in this picture. Force on the proton would have the same magnitude because the charge on a proton has the same magnitude as the charge on an electron and so the force is gonna be just the opposite direction. So 4.80 times 10 to the minus 17 newtons is the same as that on an electron but in the opposite direction to the east instead of to the west.

Comments

Why was the end part multiplied by 1/sqrt(2) ????

Thanks for the question. Each of the four charges has only a component of their electric field directed upward. That component is sin⁡(45∘)\sin(45^\circ) times kqr\dfrac{kq}{r}, where rr is the distance to the center. The field of each charge needs to by multiplied by sin⁡(45∘)\sin(45^\circ), which is a special expression that can be written in exact form as 12\dfrac{1}{\sqrt{2}}. sin⁡45=12\sin{45} = \dfrac{1}{\sqrt{2}}, is the short answer.

Hi mesmes, L/2 is the vertical component of the position, but not of the electric field. The field needs to be calculated first with the full distance from the charge, and then, as a second step, find the component of that result. It's important to do things in this order since the electric field is non-linear with distance, which is to say that when half as far, the field will not be twice as much (it'll be 4 times as much, in fact). It would be a good exercise to do the calculations with L/2, and see that the result is different.

I still don't get why the sin 45 is necessary. Why can't we just add up all the electric fields (or x4), using half the diagonal for r (.035m)?

Hello physics2aand2b, what you're describing would be fine if each electric field was pointing in the same direction. In this case, however, two of the fields are somewhat to the right, whereas the other two are somewhat to the left, and they need to be added as vectors. The symmetry gives us a shortcut where we can say "the horizontal components cancel, so let's ignore them". This leaves us with adding each of the vertical components. We can add each vertical component together since they're all pointing in the same direction (up!), and multiplying by sin 45 gives us that vertical component.
Hope this helps,
Shaun