An insulating rod having linear charge density lambda = 40.0 mu C/m and linear m

Question

An insulating rod having linear charge density lambda = 40.0 mu C/m and linear mass density mu = 0.100 kg/m is released from rest in a uniform electric field E= 100 V/m directed perpendicular to the rod (see the figure below). If L is the length of the rod, derive an expression for the force exerted by the electric field on the rod in terms of L, E, and A. What direction will the rod move? Derive the work done by the electric field after it has traveled a distance d in terms of L, E, lambda , and d Determine the speed of the rod after it has traveled 2.00 m (i.e. d = 2.00 m).

Explanation / Answer

total charge on the rod = lambda*L

(a)

force acting on the rod = F = E*Q = E*lambda*L

the rod moves to the right ( i.e in the direction of E )

(b)

work done = W = F*d = E*lambda*L*d

(c

mass = m = u*L

from work energy theorem

W = KE = 0.5*m*v^2

E*lambda*L*d = 0.5*u*L*v^2

v = sqrt((2*E*d)/u)

v = sqrt(2*100*2/0.1)

v = 63.2 m/s <<<-----answer

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