Imagine a system (triangle) consisting of three charges, q 1 = +5.00C, q 2 =+5.0
ID: 1666436 • Letter: I
Question
Imagine a system (triangle) consisting of three charges,q1 = +5.00C, q2 =+5.00C, and q3 =-5.00C, at the vertices of an equilateral trangle of sided = 2.95 cm. (a) Find the magnitude ofthe electric field at a point halfway between the chargesq1 and q2.(b) Is the magnitude of the electric field halfwaybetween the charges q2 andq3 greater than, less than, or the same as theelectric field found in part (a)? Explain. (c)Find the magnitude of the electric field at the point specified inpart (b). Imagine a system (triangle) consisting of three charges,q1 = +5.00C, q2 =+5.00C, and q3 =
-5.00C, at the vertices of an equilateral trangle of sided = 2.95 cm. (a) Find the magnitude ofthe electric field at a point halfway between the chargesq1 and q2.(b) Is the magnitude of the electric field halfwaybetween the charges q2 andq3 greater than, less than, or the same as theelectric field found in part (a)? Explain. (c)Find the magnitude of the electric field at the point specified inpart (b).
Explanation / Answer
E = -qk/r^2
Just convert all distances into x and y components and solve for Exand Ey separately. Then use Ex^2 +Ey^2 = E^2
Ex = -qk/x^2 Be sure to account for direction (i.e. if the charge is to the leftof the point in question, you multiply by an additional negativevalue to represent the direction vector being negative.)
So, for example in the x direction, we have:
Ex= -k( q1/(d/2)^2 - q2/(d/2)^2+q3/x3^2): Notice that because one of the charges (in myarbitrary example, that's q2) is to the left of center, there is anegative in front of it accounting for the direction vector. Alsonote that x3 will, in this case be equal to zero,so the q3component of the E field is equal to zero
To find the direction, just use arctan(Ey/Ex) =