Question

A moving particle encounters an external electric field that decreases its kinetic energy from 9970 eV to 6340 eV as the particle moves from position A to position B The electric potential at A is 55.0 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

q = -1.61x10⁻¹⁷ C

Step-by-step explanation:

The charge of the particle can be found using the definition of the work done by electric force:

`W = -q\Delta V` (1)

Where:

q: is the charge

ΔV: is the difference in electric potential

The work is also equal to:

`W = E_{p_(A)} - E_{p_(B)}` (2)

Where:

`E_{p_(A)}` and
`E_{p_(B)}` are the electric potential energy of the points A and B, respectively.

Now, by conservation of energy we have:

`K_(A) + E_{p_(A)} = K_(B) + E_{p_(B)}` (3)

Where:

`K_(A)` and
`K_(B)` are the kinetic energy of the points A and B, respectively.

Rearranging equation (3):

`K_(B) - K_(A) = E_{p_(A)} - E_{p_(B)}`

`K_(B) - K_(A) = W`

`K_(B) - K_(A) = -q\Delta V`

Solving the above equation for q:

`q = -(K_(B) - K_(A))/(V_(B) - V_(A)) = -(6340 eV - 9970 eV)/(19.0 V - 55.0 V) = -100.83 e \cdot (1.6 \cdot 10^(-19) C)/(1 e) = -1.61 \cdot 10^(-17) C`

Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.

I hope it helps you!