Question

An object having a mass of 11.0 g and a charge of 8.00 ✕ 10-5 C is placed in an electric field E with Ex = 5.70 ✕ 103 N/C, Ey = 380 N/C, and Ez = 0. What is the force acting on the object in each direction?

Answer 1

Answer :
`F_(x) = 45.6*10^(-2)\ N`,
`F_(y) = 3040*10^(-5) N` and
`F_(z) = 0\ N`

Explanation :

Given that,

Charge of the object q =
`8.00* 10^(-5)\ C`

Electric field in x-direction
`E_(x) = 5.70*10^(3)\ N/C`

Electric field in y- direction
`E_(y) = 380\ N/C`

Electric field in z - direction
`E_(z) = 0`

Now, using formula

`F = q E`

Now, the force on the object in x- direction

`F_(x) = 8.00*10^(-5) C *5.70*10^(3) N/C`

`F_(x) = 45.6*10^(-2)\ N`

The force on the object in y- direction

`F_(y) = 8.00*10^(-5)\ C* 380\ N/C`

`F_(y) = 3040*10^(-5) N`

The force on the object in z- direction

`F_(z) = 8.00*10^(-5)\ C* 0`

`F_(z) = 0\ N`

Hence, this is the required solution.