Question

in a region of uniform electric field as electron travels from A to B it slows from UA is equal to 6.1 into 10 to the power 6 metre per second to u b is equal to 4.5 into 10 to the power 6 metre per second what is its potential change Delta V is equal to V B minus in volts​

Answer 1

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