Question

Q/C An insulating rod having linear charge density λ=40.0σC/m and linear mass density σ = 0.100 kg/m is released from rest in a uniform electric field E=100 V/m directed perpendicular to the rod (Fig. P25.9). (a) Determine the speed of the rod after it has traveled 2.00m.

Answer 1

The speed of the rod after it has traveled 2.00m is 0.4 m/s.

How to find the speed of the rod?

Using the Law of Conservation of energy,

∆K + ∆U = 0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

1/2 mv² + (q ∆V) = 0 ............(1)

Here, m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Now, q = λL, m = L and V = Ed

Here, L is the length of the rod

E is the electric potential

d is the distance between the charges

Substituting the value of m and q in the equation 1, we get:

$(\frac{1}{2} (\mu L)v^2)-\lambda L(Ed)=0$

Solving for v, we get:

$v=\sqrt{\frac{2Ed\lambda}{\mu} }=\frac{2\times100\times40\times10^{-6}\times 2}{0.1}$

= 0.4m/sec

Hence, the speed of the rod after it has traveled 2.00m is 0.4 m/s.

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

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