Answer : [tex]F_{x} = 45.6\times10^{-2}\ N[/tex], [tex]F_{y} = 3040\times10^{-5} N[/tex] and [tex]F_{z} = 0\ N[/tex]
Explanation :
Given that,
Charge of the object q = [tex]8.00\times 10^{-5}\ C[/tex]
Electric field in x-direction [tex]E_{x} = 5.70\times10^{3}\ N/C[/tex]
Electric field in y- direction [tex]E_{y} = 380\ N/C[/tex]
Electric field in z - direction [tex]E_{z} = 0[/tex]
Now, using formula
[tex]F = q E[/tex]
Now, the force on the object in x- direction
[tex]F_{x} = 8.00\times10^{-5} C \times5.70\times10^{3} N/C[/tex]
[tex]F_{x} = 45.6\times10^{-2}\ N[/tex]
The force on the object in y- direction
[tex]F_{y} = 8.00\times10^{-5}\ C\times 380\ N/C[/tex]
[tex]F_{y} = 3040\times10^{-5} N[/tex]
The force on the object in z- direction
[tex]F_{z} = 8.00\times10^{-5}\ C\times 0[/tex]
[tex]F_{z} = 0\ N[/tex]
Hence, this is the required solution.
