Question

In: Physics

A point charge of is at the origin, and a second point charge of +6.00nC is...

A point charge of is at the origin, and a second
point charge of +6.00nC is on the x axis at x=0.800m Find
the magnitude and direction of the electric field at each
of the following points on the x axis

a)x=20.0cm b)x=1.20m c) -20.0cm

so, since its electric field question using e=k*q/r^2 i can solve this

so, e= (k*q/r^2) + (k*q/r^2)

so, +6.00nC is on the x axis at x=0.800m....

and -4.00nC is at the origin.. so 0.

a) e= (k*4*10^-9c / 2*10^-2m) + (k*6*10^-9c / .800m ?)   

it say its on the x axis at .800.. right?

Solutions

Expert Solution

E (net) = E1 + E2 + E3 + ...............


E = K q / r^2

Where K = 1 / (4 * (pi) * eo) = 9.00 * 10^9

Let the - 4.00 nC charge = - q1 and the + 6.00 nC = +q2
Let the distance from q1 to the required point = r1
Let the distance from q2 to the required point = r2

So,
a) Here,
r1 = + 20 cm =  0.2 m
r2 = 80 -20 = 60 cm =  0.6 m

E(q1) = - 9.00 * 10^9 * 4.00 * 10^(-9) / (0.2)^2 = - 900 J

E(q2) = 9.00 * 10^9 * 6.00 * 10^(-9) / (0.6)^2 = + 150 J

E(net) = (-900) + 150 = - 750 J [ the negative sign means it points in the negative x- direction]

b)
r1 = 1.2 m
r2 = (1.2 - 0.8) = 0.4 m

E(q1) = - 9.00 * 10^9 * 4.00 * 10^(-9) / (1.2)^2 = - 25 J

E(q2) = 9.00 * 10^9 * 6.00 * 10^(-9) / (0.4)^2 = 337.5 J

E(net) = 337.5 - 25 = 312.5 J [ its positive, so points along the positive x-direction]

c)
r1 = (0.2 + 0.2) = 0.4 m
r2 = (0.8 + 0.2) = 1.0 m

E(q1) = - 9.00 * 10^9 * 4.00 * 10^(-9) / (0.4)^2 = - 225 J

E(q2) = 9.00 * 10^9 * 6.00 * 10^(-9) / ( 1.0)^2 = 54 J

E(net) = - 225 + 54 = -171 J [pointing along the negative x-axis]


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