Answer:
Explanation:
Don't know what you're comparing to, but most likely the answer is the same amount of time. Mass does not affect trajectories.
Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).
The point then lies on the y-axes at d = 0.03 m.
from symmetry, the field at that point will be ascending along the y-axes.
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.
All in all, the infinitesimal field strength from the charge between x and x+dx is:
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)
Therefore, upon integration,
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) )
But lambda L = Q, the total charge on the rod, so
E = k Q / ( d * sqrt((L/2)^2 + d^2) )
Answer:
(d/λ) = 17.67
Explanation:
Equation for double-slit:
d*sinθ = m*λ
Rearrange the equation
(d/λ) = m/(sinθ)
Since it is at the second dark fringe, m = 2
Therefore:
(d/λ) = 2/(sin(6.5)
(d/λ) = 17.67
Answer:
thanks for the points man
Can someone pls help us with this question I need the answer too
