Answer Both the charges are same in magnitude and apposite in nature. So electric field at midway between charges will be in same direction and equal in magnitude. So "E_{net}=\\frac{Kq}{r^2}+\\frac{Kq}{r^2}" "E_{net}=\\frac{2Kq}{r^2}" "E_{net}=\\frac{2(9\\times10^9)(2\\times10^{-7})}{(0.5)^2}=144\\times10^{2}N"
point A is between two charges, one to the left (+2 microcoul) and one to the right (-2 microcoul). point A is midway between the two charges. the two charges are 5cm apart. what is the electric field in the plane of the page (diagram) at a point 5cm left from point A. give in terms of x and y components of the electric field assume positive x is directed to the right, and positive y is directed upward of point A
my solution is incorrect. what did i do wrong? did i interpret the question correctly (plane part), if so could you explain what the "plane" part means? Answers and Replies
no, the positive charge in the diagram is on the left, not the right. how would having the charge on either side affect the result?
It changes the direction of the field. I think the problem is that you are using the formula somewhat blindly to calculate the field and its sign. Instead, use the formula to get the magnitude of the field, then figure out the direction based on the sign of the charge. The field always points away from a positive charge and toward a negative charge.
i can't see how i got the charges wrong in my calculations...q_l = q_left = left charge = positive charge = +2*10^-6 coulombs, and the distance is 2.5cm from the positive charge, so 0.025m. as for the q_r = q_right = negative charge = -2*10^-6 coulombs, and the distance is 7.5cm from the negative charge, so 0.075m
am i not seeing something?
Which way does the field from that negative charge point? Should the sign of its field be positive or negative?
the positive charge electric field points away from the charge, it should be negative the negative charge electric field points into the charge, it should be positive
is i change the signs in my original calculations, will my answer be correct?
thanks so much , it worked!! it was the sign all along, that's something i need to remember
just another quick question. if was to determine the electric field at point P, is this how i would do it. E1 = kq_l/r_1^2
= (9*10^9)(-2*10^-6) / (0.025^2)
= -2.88*10^7 n/c E2 = kq_r/r_2^2
= (9*10^9)(+2*10^-6) / (0.025^2)
= +2.88*10^7 n/c so Enet = E1 + E2 = 0 so (Ex, Ey = 0, 0) correct? apparently it is wrong.
i know there is no field in the y direction so it has to be zero, and the distance from either charge is equal and only along the x axis, and since they are opposite directions, shouldn't they cancel out to zero?
(I assume you're talking about point A, right in the middle between the two charges.)
i thought my signs were correct this time? i made them such that they were in the same directions as the first problem, after we corrected them. if i reverse the signs, the magnitude will be same, and wouldn't they just cancel out? and yes i was talking about point A, in the middle when is said 'opposite directions' i meant away from the positive, towards the negative. or is that incorrect wording? the positive points away towards the negative, and the electric field of the negative flows inwards to itself
so the positive charge, sign is negative, and the negative charge, the sign is positive. how would that not cancel out if you get same magnitude?
Actually, if you stuck to that last sentence you'd be OK.  While your previous sentence was OK, this conclusion is false. (Actually draw the field arrows on a diagram.)
since the point is now between the charges, the positive points to the right so it is positive, and the negative charge's field is positive direction too
(If both charges where the same sign, then the field in the middle would be zero.) |
In the figure, point p is midway between the two charges.
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