Electrons are emitted by a hot filament and are accelerated by an electric field, as shown in the figure (32-Q1). The two stops at the left ensure that the electron beam has a uniform cross-section.

(a) The speed of the electrons is more at B than at A.
(b) The electric current is from left to right.
(c) The magnitude of the current is larger at B than at A.
(d) The current density is more at B than at A.

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Solution

(a) The speed of the electrons is more at B than at A.

Let the potentials at A and B be V_A and V_B.
As potential, $E = - \frac{d V}{d r}$ ,
potential increases in the direction opposite to the direction of the electric field.
Thus, V_A < V_B
Potential energy of the electrons at points A and B:
U_A= $-$ eV_A
U_B= $-$ eV_B
Thus, U_A> U_B
Let the kinetic energy of an electron at points A and B be K_Aand K_B respectively.

Applying the principle of conservation of mechanical energy, we get:

U_A+ K_A = U_B+ K_B
As, U_A> U_B,
K_A < K_B
Therefore, the speed of the electrons is more at B than at A.

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Similar questions

A

B

Table 1	Table 2
(a) current at A	(p) is zero
(b) drift velocity of electrons at A	(q) is more than at B
(c) electric field in the wire at A	(r) is less than at B
(d) current density at A	(s) is equal to that at B

I = n e A V_{d}

V_{d}

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