Question

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 1

Answer:

Electric force, $F=2.88\times 10^{-14}\ N$

Explanation:

It is given that,

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

Charge, $q=4.8\times 10^{-19}\ C$

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 :

$F=q\times E$

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

$F=2.88\times 10^{-14}\ N$

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