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djverab [1.8K]
3 years ago
9

A positive charge is placed at rest at the center of a region of space in which there is a uniform electric field. (A uniform fi

eld is one whose strength and direction are the same at all points within the region.) What happens to the electric potential energy of the positive charge, after the charge is released from rest in the uniform electric field?A. It will remain constant because the electric field is uniform.B.It will decrease because the charge will move in the opposite direction of the electric field.C.It will remain constant because the charge remains at rest.D.It will decrease because the charge will move in the direction of the electric field.E.It will increase because the charge will move in the direction of the electric field.
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

.D.It will decrease because the charge will move in the direction of the electric field

Explanation:

<em>which there is a uniform electric field. (A uniform field is one whose strength and direction are the same at all points within the region.) What happens to the electric potential energy of the positive charge, after the charge is released from rest in the uniform electric field?A. It will remain constant because the electric field is uniform.B.It will decrease because the charge will move in the opposite direction of the electric field.C.It will remain constant because the charge remains at rest.D.It will decrease because the charge will move in the direction of the electric field.E.It will increase because the charge will move in the direction of the electric field.</em>

solution

The potential energy decreases, converted to kinetic energy

The charge will feel a force in the direction of the electric field (F=Eq) and thus it will accelerate with

a constant acceleration. (Just like releasing an object above the earth's surface - constant acceleration,

at least until it hits something.)

tus te answer will be

.D.It will decrease because the charge will move in the direction of the electric field

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Explanation:

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\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3

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F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N

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Answer:

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Explanation:

To solve this problem, we use the right motion equation.

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The right motion equation is;

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