Question

In a region of uniform electric field,as an electron travels from A to B,it slows from Ua=10 6m/s to Ub=4.5×10 6 m/s.What is its potential change ∆V=Vb-Va in volts?

Answer 1

Answer:

Explanation:

=> It is given here,

$M_{e}$ = 9.1 * $10^{-31}$ Kg

$q_{e}$ = 1.6 * $10^{-19}$

$u_{A}$ = 6.1 * $10^{6}$ m/s

$u_{B}$ = 4.5 * $10^{6}$ m/s

=> W.D = ΔKE

= 1/2 n $(u_{B}^2 - u_{A}^2)$

= 1/2 * 9.1 * $10^{-31} (4.5^2 - 6.1^2) * 10^{12}$

= $\frac{9.1}{2} * (20.25 - 37.21)* 10^{-31+12}$

= $\frac{9.1}{2} * (-16.96)* 10^{-19}$

=> W.D = $q (V_B - V_A)$

$(V_B - V_A)$ = W.D / q

= $\frac{-9.1 * 16.96 * 10^{-19}} {-1.6 * 10^{-19}*2}$

=$\frac{9646}{2}$

=48.23

$V_{B} - V_{A}$ = + 48 V

Thus, In a region of uniform electric field,as an electron travels from A to B, its potential change is + 48 V.

Answer 2

change in kinetic energy = $\frac{1}{2}m(v_B^2-v_A^2)$

where m is mass of electron , $v_A$ is velocity of electron at A and $v_B$ is velocity of particle at B.

we know, mass of electron , m = 9.1 × 10^-31 Kg, $v_A$ = 6.1 × 10^6 m/s and $v_B$ = 4.5 × 10^6 m/s

so, change in kinetic energy = 1/2 × 9.1 × 10^-31 [(4.5 × 10^6 )² - (6.1 × 10^6)²]

= 4.55 × 10^-31 × 10¹² [ (4.5 +6.1)(4.5-6.1)]

= -77.168 × 10^-19 J

= -7.7168 × 10^-18 J

now, change in kinetic energy = workdone by electrical potential

or, -7.7168 × 10^-18 = $q(V_B-V_A)$ where q is charge of electron,

so, q = -1.6 × 10^-19 C

now, -7.7168 × 10^-18 = 1.6 × 10^-19 $(V_B-V_A)$

or, $(V_B-V_A)$ = 77.168/1.6

= 48.23 Volts