Question

An electron moving parallel to the x-axis has an initial speed of 4.98 × 10⁶m/s at the origin. Its speed is reduced to 1.76 × 10⁵m/s at the point x= 2.00 cm.

(a) Calculate the electric potential difference between the origin and that point.

(b) Which point is at the higher potential?

Answer 1

Answer 2

conservation of energy

0.5 m v1^2 - eV1 = 0.5 m v2^2 -e V2

where m = mass of electron

v1 intial speed = 4.98*10^6 m/s

v2 final speed = 1.76*10^5 m/s

e = charge of electron

for b

the point where V is higher

Answer 3

CONSERVATION OF ENERGY

0.5mv1^2-ev1=0.5 mv2^2-ev2

where m=mass of electron

v1is initial speed=4.8*10^6m/s

v2is final speed=1.76*10^5m/s

e=charge of electron

for b

the point v is higher