In: Physics
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?
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]