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Lerok [7]
2 years ago
7

Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

e right hand and has magnetic field of 4.00 N/C .If an ion of charge +2e is released at rest at the left hand plate .What its K.E when reaches the right hand plate?tell the answer with expalanation
Physics
1 answer:
dezoksy [38]2 years ago
4 0
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>

Explanation:

When the charge of 2e is placed in between the plates .

The force applied on this charge by plates is = q E

here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C

and E is the magnitude of electric field intensity

The work done = Force x displacement

Thus W = q E x S

here S is displacement

Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8

= 1.02 x 10⁻¹⁷ J

This work will be converted into the kinetic energy of charge .

Thus K.E = 1.02 x 10⁻¹⁷ J

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Answer:

0.30581

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Explanation:

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F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N

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F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581

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The coefficient of kinetic friction is 0.24464

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