- #1
symmet
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Here's the problem:
Three charges are at the corners of an equilateral triangle as in the figure below:
....(+)q_1
.../...\
.../...\ L
.../...\
...q_0(+)____(-)q_2L = .06 m
q_0 = 1.8*10^-6 C
q_1 = 6.6*10^-6 C
q_2 = -4.6*10^-6 C
Calculate the electric field at the position of q_0 (the origin) due to q_1 and q_2.
Answer is a vector in terms of i and j.
My approach was to break down q_1 and q_2 into vector components. For example, to find the horizontal components, I did:
E_x = k(q/x^2)
where x is the horizontal distance from position q_0 and q is the charge. (k = 8.99*10^9)
So, plugging this in for both q_1 and q_2, I got:
q_1: E_x = -659266.67 N/C
q_2: E_x = 114872.22 N/CI put a - on q_1 because of the direction of the vector at q_0. Adding these together I get -544394.45 for the total horizontal component. But this isn't right. Can anyone tell me where I've gone wrong in my thought process? Thanks.
Three charges are at the corners of an equilateral triangle as in the figure below:
....(+)q_1
.../...\
.../...\ L
.../...\
...q_0(+)____(-)q_2L = .06 m
q_0 = 1.8*10^-6 C
q_1 = 6.6*10^-6 C
q_2 = -4.6*10^-6 C
Calculate the electric field at the position of q_0 (the origin) due to q_1 and q_2.
Answer is a vector in terms of i and j.
My approach was to break down q_1 and q_2 into vector components. For example, to find the horizontal components, I did:
E_x = k(q/x^2)
where x is the horizontal distance from position q_0 and q is the charge. (k = 8.99*10^9)
So, plugging this in for both q_1 and q_2, I got:
q_1: E_x = -659266.67 N/C
q_2: E_x = 114872.22 N/CI put a - on q_1 because of the direction of the vector at q_0. Adding these together I get -544394.45 for the total horizontal component. But this isn't right. Can anyone tell me where I've gone wrong in my thought process? Thanks.
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