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Question
Electrons are emitted by a hot filament and are accelerated by an electric field, as shown in the figure. The two stops at the left ensure that the electron beam has a uniform cross-section.
Options
The speed of the electrons is more at B than at A
The electric current is from left to right
The magnitude of the current is larger at B than at A
The current density is more at B than at A
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Solution
The speed of the electrons is more at B than at A
Let the potentials at A and B be VA and VB.
As potential,
\[E = - \frac{dV}{dr}\]
potential increases in the direction opposite to the direction of the electric field.
Thus, VA < VB
Potential energy of the electrons at points A and B:-
UA = -eVA
UB = -eVB
Thus, UA > UB
Let the kinetic energy of an electron at points A and B be KA and KB respectively.
Applying the principle of conservation of mechanical energy, we get:-
UA + KA = UB + KB
As, UA > UB,
KA < KB
Therefore, the speed of the electrons is more at B than at A.
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