A point charge q_1 = -2.0 C is at the origin, and a second point charge q_2 = +4

Question

A point charge q_1 = -2.0 C is at the origin, and a second point charge q_2 = +4.8 C is at the point x = 1.1 m, y = 2.9 m. (a) Find the x and y coordinates of the position at which an electron would be in equilibrium. x = y = (b) Find the x and y coordinates of the position at which a proton would be in equilibrium. x = y = (c) Suppose the charges q_1 and q_2 were both doubled. How would your answers change? The electron would be in equilibrium at the following point. x = y = The proton would be in equilibrium at the following point. x = y = Why?

Explanation / Answer

given, q1 = -2 C, at origin

q2 = 4.8 C, at x = 1.1m , y = 2.9 m

a. as both the charges are of different sign

equilibrium can be achieved at either side of both the charges and not in between the two

also the point of consern should lie on the straight line joining the two points

so consider just the two charges

distance between two, r = sqroot(1.1^2 + 2.9^2) = 3.101 m

let distance of the equilibrium point be d from q1

so, electric field at this point

E = kq1/d^2 + kq2/(d + r)^2 = 0

2/d^2 = 4.8/(d + r)^2

(d + r)^2 = 2.4d^2

d^2 + r^2 +2dr = 2.4d^2

1.4d^2 - 2dr - r^2 = 0

1.4d^2 - 2*D*3.101 - 3.101^2=0

1.4d^2 - 6.202d - 9.62 = 0

solving for d

d = 5.646m

so, coordinates of the point of interest

x = -dcos(theta)

y = -dsin(theta)

where tan(thjeta) = 2.9/1.1

theta = 69.22 deg

so (x,y) = (-2.003, -5.27)

b. as the electric field is 0 at this point, no matter if a proton or an electron is used as a test chyarge, they all will be at equilibrium at this point

c. If the charges were doubled, the answer would still remain the same as chamnging both charges by same factor cancelsd out the effect of this change at the point of equilibrium

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