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

Electric force,
`F=2.88* 10^(-14)\ N`

Step-by-step explanation:

It is given that,

Magnitude of electric field,
`E=6* 10^4\ N/C`

Charge,
`q=4.8* 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* E`

`F=4.8* 10^(-19)\ C* 6* 10^4\ N/C`

`F=2.88* 10^(-14)\ N`

So, the electric force on the charge is
`2.88* 10^(-14)\ N`. Hence, this is the required solution.