Answer:
Charge, [tex]q=3.24\times 10^{-18}\ C[/tex]
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 :
[tex]\Delta KE=-q(V_B-V_A)[/tex]
[tex]q=\dfrac{\Delta KE}{(V_B-V_A)}[/tex]
[tex]q=\dfrac{ 9650\ eV-8900\ eV}{(19\ V-56\ V)}[/tex]
q = -20.27 e
[tex]q =-20.27e\times \dfrac{1.6\times 10^{-19}\ C}{e}[/tex]
[tex]q=-3.24\times 10^{-18}\ C[/tex]
Hence, this is the required solution.
