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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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<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

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For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

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We are given:

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238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

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