Question

91+2.60 x 10-6 C, q2 = +3.75 x 10-6 C, and93 +1.30 x 10-6 C. Find the x-component ofthe net force on q₁. Include the correct + or -sign to indicate direction.91=0.283 m0.200 m9245.0°0.200 m93(Make sure you know the direction of each force!)

Answer 1

The x-component of the net force on
`\( q_1 \)` is approximately
`\(-0.0142\)` N (rounded to four decimal places), with the negative sign indicating that the net force is in the negative x-direction.

To find the x-component of the net force on
`\( q_1 \)` due to the charges
`\( q_2 \)` and
`\( q_3 \)`, we can use Coulomb's law, which states that the force between two charges is:

`\[ F = k (|q_1 q_2|)/(r^2) \]`

where:

- F is the magnitude of the force between the charges,

- k is Coulomb's constant
`(\( 8.988 * 10^9 \) N m²/C²),`

-
`\( q_1 \)` and
`\( q_2 \)` are the magnitudes of the charges,

- r is the distance between the charges.

The x-component of the force can be found by multiplying the force by the cosine of the angle with respect to the x-axis.

We have three charges:

-
`\( q_1 = +2.60 * 10^(-6) \) C,`

-
`\( q_2 = +3.75 * 10^(-6) \) C,`

-
`\( q_3 = +1.30 * 10^(-6) \) C.`

The distances are:

- Between
`\( q_1 \)` and
`\( q_2 \)`:
`\( r_(12) = 0.283 \)` m,

- Between
`\( q_1 \)` and
`\( q_3 \): \( r_(13) = 0.200 \)` m (since
`\( q_3 \)` is directly on the x-axis from
`\( q_1 \)`, the force between
`\( q_1 \)`and
`\( q_3 \)`will only have an x-component).

The angle between
`\( q_1 \)`and
`\( q_2 \) is \( 45^\circ \)`, which means the x-component of the force due to q2 will involve the cosine of
`\( 45^\circ \).`

Let's calculate the x-component of the force on q1 due to q2 first, and then add the force due to q3 (which lies on the x-axis, so its entire force is an x-component).

The force from q2 on q1 is repulsive and to the left (negative x-direction), and the force from q3 on q1 is also repulsive and to the right (positive x-direction) because like charges repel.

Answer 2

Using coulomb's law:

`\begin{gathered} F_(21)=K\cdot(q_1\cdot q_2)/(r^2) \\ F_(21)=(8.988*10^9)\cdot((2.6*10^(-6))(3.75*10^(-6)))/(0.283^2) \\ F_(21)=1.094N \end{gathered}`
`\begin{gathered} F_(31)=K\cdot(q_1\cdot q_3)/(r^2) \\ F_(31)=(8.988*10^9)\cdot((2.6*10^(-6))(1.3*10^(-6)))/(0.2^2) \\ F_(31)=0.759N \end{gathered}`

Now, let's calculate the net force on each component:

`\begin{gathered} F_x=F_(21)cos(45)=0.774N \\ F_y=F_(31)+F_(21)sin(45)=1.533N \end{gathered}`

Now, we can calculate the net force and the direction:

`\begin{gathered} F_(net)=√(F_x^2+F_y^2) \\ F_(net)=√((0.774^2)+(1.533^2)) \\ F_(net)=+1.717N \\ \end{gathered}`

Answer:

+ 1.717 N