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The electric field at a particular location is measured to be N/C. what force would a alpha particle experience if placed at this particular location? F=_______.

Answer :

lublana

Answer:

[tex]<0,-8.32\times 10^{-17},0>N[/tex]

Explanation:

We are given that

Electric field at a particular location=[tex]<0,-260,0>N/C[/tex]

Charge on alpha particle,q=2e

Charge on alpha particle,q=[tex]2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C[/tex]

Where 1 e=[tex]1.6\times 10^{-19}C[/tex]

Force on alpha particle=qE

Where q= Charge on  particle

E=Electric field

Using the formula

Force on alpha particle=[tex]3.2\times<0,-260,0>[/tex]

Force on alpha particle=[tex]<0,-8.32\times10^{-17},0>N[/tex]

Hence, the force at this particular location=[tex]<0,-8.32\times 10^{-17},0>N[/tex]

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