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vlabodo [156]
4 years ago
8

A moving particle encounters an external electric field that decreases its kinetic energy from 9650 eV to 8900 eV as the particl

e 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.
Physics
1 answer:
mafiozo [28]4 years ago
6 0

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.

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