415V 505V 515V
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There are 3 points w/same distance from each other (6x10-3m between each point). What is the magnitude and direction of electric field (E) in the vicinity of the origin (the middle point – 505V) ?
Thank you!
To find the magnitude and direction of the electric field (E) in the vicinity of the origin, we can use the concept of superposition.
First, let's consider the electric field due to the 415V point. The electric field (E1) at the origin due to this point can be calculated using Coulomb's law:
E1 = k * Q1 / r1^2
where k is the electrostatic constant (approximately 9 × 10^9 N m^2/C^2), Q1 is the charge at the 415V point, and r1 is the distance from the 415V point to the origin (which is half the distance between the 415V and 505V points, so r1 = 3x10^-3 m).
Similarly, we can calculate the electric field (E2) at the origin due to the 515V point:
E2 = k * Q2 / r2^2
where Q2 is the charge at the 515V point and r2 is the same as r1.
Since the points are equidistant from each other and the electric fields due to these two points have the same magnitude, we can assume that the electric field due to the 505V point is in the direction of the origin and has the same magnitude as the other two electric fields (E1 and E2).
Therefore, the total electric field (E) at the origin is the vector sum of E1, E2, and the electric field due to the 505V point. Since the electric fields due to the 415V and 515V points are in opposite directions, they will cancel each other out, leaving only the electric field due to the 505V point:
E = E1 + E2
Finally, we can calculate the magnitude and direction of the electric field at the origin by substituting the values of E1 and E2 into the equation for E, and taking into account the direction:
E = E1 + E2 = k * Q1 / r1^2 + k * Q2 / r2^2
where E1 and E2 have the same magnitude, and their directions are opposite. Note that the direction of E will be towards the 505V point.