Now find the gravitational potential energy U ( z ) of the object when it is at

Question

Now find the gravitational potential energy U(z) of the object when it is at an arbitrary height z. Take zero potential to be at position z=0. Keep in mind that the potential energy is a scalar, not a vector.

Express U(z) in terms of m, z, and g.

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The correct answer does not depend on: z^.

Part C

In what direction does the object accelerate when released with initial velocity upward?

In what direction does the object accelerate when released with initial velocity upward?

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Correct

Electric Force and Potential Energy

Now consider the analogous case of a particle with charge q placed in a uniform electric field of strength E, pointing downward (in the k^ direction)

Part D

Find F (z), the electric force on the charged particle at height z.

Express F (z) in terms of q, E, z, and k^.

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Part E

This question will be shown after you complete previous question(s).

Part F

In what direction does the charged particle accelerate when released with upward initial velocity?

In what direction does the charged particle accelerate when released with upward initial velocity?

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Electric Field and Electric Potential

The electric potential V is defined by the relationship U=qV, where U is the electric potential energy of a particle with charge q.

Part G

Find the electric potential V of the uniform electric field E =Ek^. Note that this field is not pointing in the same direction as the field in the previous section of this problem. Take zero potential to be at position z=0.

Express V in terms of q, E, and z.

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Part H

The electric field can be derived from the electric potential, just as the electrostatic force can be determined from the electric potential energy. The relationship between electric field and electric potential is E = V, where   is the gradient operator:

V=Vxi^+Vyj^+Vzk^.

The partial derivative Vx means the derivative of V with respect to x, holding all other variables constant.

Consider again the electric potential V=Ez corresponding to the field E =Ek^. This potential depends on the z coordinate only, so Vx=Vy=0 and Vz=dVdz.

Find an expression for the electric field E  in terms of the derivative of V.

Express your answer as a vector in terms of the unit vectors i^, j^, and/or k^. Use dV/dz for the derivative of V with respect to z.

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Part I

A positive test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy.

Choose the appropriate answer combination to fill in the blanks correctly.

Choose the appropriate answer combination to fill in the blanks correctly.

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Part J

A negative test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy.

Choose the appropriate answer combination to fill in the blanks correctly.

Choose the appropriate answer combination to fill in the blanks correctly.

Explanation / Answer

B. U(z) = U(z) - U(0) = mgz
C. As the force applicable on the object is in downwards direction, the object accelerated in the downwards direction
D. F = -qEk^
F. upward or downward depending on its charge
G. V = -integral[E.dz] = -integral[Ek^ . dzk^] = -integral[Edz] = -E[z]
H. Given V = -Ez
differentiating both sides wrt z
dV/dz k = -E
E = -dV/dz k
I. Positive test charge will accelerate towards region of lower PE and lower potential
J. Negative test charge will accelerate towards region of higher PE and higher potential

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