Three equal positive charges q are at the corners of an...

Three equal positive charges q are at the corners of an equilateral triangle of side a as shown in the figure below. Assume the three charges together create an electric field. P, q, a, a, a, q, q, i (b) Find the location of one point (other than ?) where the electric field is zero. at P at the center of the triangle at each vertex at the center of the base (c) What is the magnitude of the electric field at P due to the two charges at the base? (Use any variable or symbol stated above along with the following as necessary: ke.) E = ?3 k e q / a^2

Transcript

00:01 Hello, here we have an equilateral triangle and at the vertices of this triangle, three charges are placed.

00:17 So, in each vertex there is a charge of q and each side is a.

00:27 First of all, we have to determine at which field, at which point the field is zero.

00:38 So, and you are right that is in the center of this triangle.

00:41 So it means that we have to draw the median perpendiculars to each side so that's kind of not the best drawing but this point h is where the field is zero because the system is symmetric and it's symmetric with respect to point h so now let's answer questions to do so to do this we to find the field at point p we have to introduce x and y axis so why we'll direct y upwards and will direct x upwards x upwards x to the right both charges are positive so therefore let's label points as one and two and let's show e1 and e2 e2 e1 and e2 e1 and 2 have the same absolute value absolutely has the same magnitude but different direction and as we can notice e1 x equals to negative e2 x therefore e at point p equals to e1y plus e2y which is 2 e1y now we have to find e1y e1y e 1y equals to k k k k k e squared over sorry k q u over a squared times times sign of angle alpha and now let's find this angle alpha how would we do this so let's prolong x -axis and let's look at let's find this angle alpha so as we can notice angle alpha is also here and alpha equals to 90 degree minus this angle which is 60 degree over 2 because again that's an equilateral triangle so that equals to 60 degree then e at point p equals to 2 kq squared over a squared times sign of 60 degree which is which is square root of 3 over 2 so that is square root of 3 times key e over q.

03:44 Oh yeah, of course there is no square...

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