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Fiesta28 [93]
3 years ago
9

g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

ve constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

F_x = -\frac{6 C_6}{2^7}

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by \frac{C_6}{X_6}

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = \frac{C_6}{X_6}

Force exerted by one atom upon another

F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X}  (-\frac{C_6}{X^6})

or

F_x = \frac{\partial}{\partial X}  (\frac{C_6}{X^6})

or

F_x = -\frac{6 C_6}{2^7}

As we can see that the C_6 comes in positive and constant which represents that the force is negative that means the force is attractive in nature

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Qx is the charge at X and Qy is the charge at Y.

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Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

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C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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