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

Answer:

q = -1.61x10⁻¹⁷ C

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:

$q = -\frac{K_{B} - K_{A}}{V_{B} - V_{A}} = -\frac{6340 eV - 9970 eV}{19.0 V - 55.0 V} = -100.83 e \cdot \frac{1.6 \cdot 10^{-19} C}{1 e} = -1.61 \cdot 10^{-17} C$                                              

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

I hope it helps you!