Four electric charges are at the corners of a square of side a. Using a standard

Question

Four electric charges are at the corners of a square of side a. Using a standard Cartesian coordinate system the charge at point x = 0, y = 0 is The remaining three charges at the other corners of the square are -q. (a) Find the magnitude and direction of the electric field at the position of the charge -q, the coordinates of which are x = a, y = a. (b) What is the electric force on this charge? A proton accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 1.20 x 10^6 m/s (nonrelativistic since v is much less than the speed of light). The mass of a proton is 1.67 x 10^-27 kg. Find the acceleration of the proton. How long does it take the proton to reach this velocity? A uniformly charged rod of length l is bent into the shape of a semicircle. If the rod has a total charge of -ft find the magnitude and direction of the electric field at 0, the center of the semicircle. Which one of the diagrams below is not a possible electric field configuration for a region of space which does not contain any charges? Why?

Explanation / Answer

(1)
Net Electric field on charge -q @ (a,a) is given by,
Enet = k*q/a^2 i^ + k*q/a^2 j^ - k*q/(2*a^2) * cos(45) i^ -  k*q/(2*a^2) * cos(45) j^
Enet = (k*q/a^2 - k*q/(2*a^2) * cos(45)) i^ + (k*q/a^2 - k*q/(2*a^2) * cos(45)) j^
Enet = k*q/a^2 *[ (1 - cos(45)/2) i^ + (1 - cos(45)/2) j^ ]
Enet = 0.646 * k*q/a^2 i^ + 0.646 * k*q/a^2 j^

|Enet | = sqrt(0.646^2 + 0.646^2)
|Enet | = 0.914 k*q/a^2 N/C

Direction = North East

(b)

We know, F= q*E
Net Electric Force on the charge, Fnet =  0.914 (k*q^2/a^2) N/C

(2)
E = 640 N/C
v = 1.2 * 10^6 m/s
m = 1.67 * 10^-27 Kg

(a)
Force on proton, F = q*E
We also know,
F = m*a
Som
m*a = q*E
1.67 * 10^-27 * a = 1.6*10^-19 * 640
a = 6.13 * 10^10 m/s^2

(b)
Using motion equation,
v = u + a*t
1.2 * 10^6 = 0 + 6.13 * 10^10 * t
t = 1.96 * 10^-5 s

please post sepearate questions in separate post only !!

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