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

Answer:

Charge, $q=3.24\times 10^{-18}\ C$

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 :

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

$q=\dfrac{\Delta KE}{(V_B-V_A)}$

$q=\dfrac{ 9650\ eV-8900\ eV}{(19\ V-56\ V)}$

q = -20.27 e

$q =-20.27e\times \dfrac{1.6\times 10^{-19}\ C}{e}$

$q=-3.24\times 10^{-18}\ C$

Hence, this is the required solution.