An electric field with a magnitude of 6.0 × 104 N/C is directed parallel to the positive y axis. A particle with a charge q = 4.8 × 10–19 C is moving along the x axis with a speed v = 3.0 × 106 m/s. The force on the charge is approximately:

Answer :

Answer:

Electric force, [tex]F=2.88\times 10^{-14}\ N[/tex]

Explanation:

It is given that,

Magnitude of electric field, [tex]E=6\times 10^4\ N/C[/tex]

Charge, [tex]q=4.8\times 10^{-19}\ C[/tex]

The electric field is directed parallel to the positive y axis. We need to find the  force on the charge particle. It is given by :

[tex]F=q\times E[/tex]

[tex]F=4.8\times 10^{-19}\ C\times 6\times 10^4\ N/C[/tex]

[tex]F=2.88\times 10^{-14}\ N[/tex]

So, the electric force on the charge is [tex]2.88\times 10^{-14}\ N[/tex]. Hence, this is the required solution.

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