An ion with charge +qis in a region of uniform electric field of magni...
Electric force, F = qE
Work done = W = F.d = qEd.
But the change is electrostatic energy = –W
So potential energy decreased.
The correct answer is: The electric potential energy of the ion decreased by the amount qEd.
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An ion with charge +qis in a region of uniform electric field of magni...
Explanation:
The electric potential energy of a charged particle in an electric field is given by the equation U = qV, where U is the electric potential energy, q is the charge of the particle, and V is the electric potential.
When the charged particle is moved a distance d in the direction of the electric field, the work done on the particle is given by the equation W = Fd, where W is the work done, F is the force on the particle, and d is the distance moved.
The force on a charged particle in an electric field is given by the equation F = qE, where F is the force, q is the charge of the particle, and E is the electric field.
From the given information, we can see that the charge q is moved a distance d in the direction of the electric field. This means that the force on the particle is in the same direction as the displacement, so the work done on the particle is positive.
Now, let's consider the equation for work done:
W = Fd
Substituting the equation for force:
W = qEd
We can see that the work done on the particle is equal to qEd.
The electric potential energy is defined as the work done on a particle to move it from infinity to its current position in an electric field. Therefore, the change in electric potential energy (ΔU) is equal to the negative of the work done on the particle:
ΔU = -W
Substituting the equation for work done:
ΔU = -qEd
Since the question asks for the change in electric potential energy, we can see that the electric potential energy decreased by the amount qEd.
Therefore, the correct answer is option A: The electric potential energy of the ion decreased by the amount qEd.