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Ksivusya [100]
3 years ago
6

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
________ × 10–4 J
What is the electric potential energy of the point charge at 1.3 m?

_________ × 10–3 J
Physics
2 answers:
schepotkina [342]3 years ago
3 0
4 for the first one and 5.2 for the second one.
natali 33 [55]3 years ago
3 0

E = electric field = 1 x 10³ N/C

q = charge = 4 x 10⁻⁶ C

d = distance = 0.10 m

electric potential energy is given as

U = q E d

inserting the above values

U = (4 x 10⁻⁶ ) (1 x 10³) (0.1)

U = 4 x 10⁻⁴ J


when distance = d = 1.3 m

electric potential energy is given as

U = q E d

inserting the above values

U = (4 x 10⁻⁶ ) (1 x 10³) (1.3)

U = 5.2 x 10⁻³ J

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A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se
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Answer:

W_2=-12J

Explanation:

The work of force 2 will be given by the vectorial equation W_2=F_2.d. We know the value of F_1 and have information about its movement, which relates to the net force F=F_1+F_2.

About this movement we can obtain the acceleration using the equation v_f^2=v_i^2+2ad. Since it departs from rest we have a=\frac{v_f^2}{2d}.

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With our values:

W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J

<em>Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:</em>

<em>W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}</em>

<em>Which leads us to the previous equation straightforwardly.</em>

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