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 : $F_{x} = 45.6\times10^{-2}\ N$, $F_{y} = 3040\times10^{-5} N$ and $F_{z} = 0\ N$

Explanation :

Given that,

Charge of the object q = $8.00\times 10^{-5}\ C$

Electric field in x-direction $E_{x} = 5.70\times10^{3}\ N/C$

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

$F_{x} = 8.00\times10^{-5} C \times5.70\times10^{3} N/C$

$F_{x} = 45.6\times10^{-2}\ N$

The force on the object in y- direction

$F_{y} = 8.00\times10^{-5}\ C\times 380\ N/C$

$F_{y} = 3040\times10^{-5} N$

The force on the object in z- direction

$F_{z} = 8.00\times10^{-5}\ C\times 0$

$F_{z} = 0\ N$

Hence, this is the required solution.