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In the figure below, determine the point (other than infinity) at which the total electric field is zero.

m to the left of ?2.5 ï¿½ 10?6 C charge

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Added Mon, 14 Dec '15
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Recall the definition of electric field: E = ke*Q/r^2. Essentially, each charge will produce its own electric field that changes with distance. You're being asked to solve for the point where these two fields will sum to zero. In other words, find the point at which E1 = -E2. To do this, we must calculate their electric fields in order to find the distance from each of these two charges. Please note that because the left charge is weaker, it's safe to assume that the location of this point is going to be to the left of the left charge. We're given that they are separated by 1m, thus d2 = d1-1, since both distances are going to be negative.

E1 = ke(-2.5uC)/d1^2

E2 = ke(6uC)/(d1-1)^2

E1 + E2 = 0 -> -2.5/d1^2 + 6/(d1-1)^2 = 0 -> 6/(d1-1)^2 = 2.5/d1^2 -> 6/2.5 = d1^2/(d1-1)^2

2.4 = d1^2/(d1-1)^2 -> 1/sqrt(2.4) = (d1-1)/d1 = 1 - 1/d1 -> 1/d1 = 1 - 1/sqrt(2.4)

-> d1 = 1/(1-(1/(sqrt(2.4)))) = 2.8209m

So now that we've calculated the value for d1, and we know that d2 = d1 - 1 = 1.8209m, let's test these values to see if they work (we just need to make sure that Q/r^2 is equal since ke is constant). Assuming that the point is to the left of q = -2.5uC, the distance from this particle is the smaller.

6/(2.8209m)^2 = 2.5/(1.8209m)^2

0.754 = 0.754

So it checks out. Thus the distance is 1.8209m left of the negative particle.

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Added Mon, 14 Dec '15
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