Question

A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particle moves from position A to position B The electric potential at A is 56.0 V and that at B is 19.0 V Determine the charge of the particle Include the algebraic sign or with your answer.

Answer 1

Charge,
`q=3.24* 10^(-18)\ C`

Step-by-step explanation:

A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particle moves from position A to position B The electric potential at A is 56.0 V and that at B is 19.0 V. We need to find the charge of the particle.

It can be calculated using conservation of energy as :

`\Delta KE=-q(V_B-V_A)`

`q=(\Delta KE)/((V_B-V_A))`

`q=( 9650\ eV-8900\ eV)/((19\ V-56\ V))`

q = -20.27 e

`q =-20.27e* (1.6* 10^(-19)\ C)/(e)`

`q=-3.24* 10^(-18)\ C`

Hence, this is the required solution.