Aparticle (charge = 50 μc) moves in a region where the only force on it is an electric force. as the particle moves 25 cm from point a to point b, its kinetic energy increases by 1.5 mj. determine the electric potential difference, vb − va.

Vb - Va = -30V

Explanation:

Given data

charge Q = 50 μC

kinetic energy E =  1.5 mJ

particle moves = 25 cm

to find out

the electric potential difference

solution

we have find potential difference VB − VA tha is Vab

we have energy point charge i.e

E =  QV

put value

1.5 × = 50 ×  × Vab

Vab = 1.5 × / 50 ×

Vab = 30 V

we know that

Vab = - Vba

so Vba = -30 V

Vb - Va = -30V

it was given as a lesson and guide for prayer.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2