Question

Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

Answer 1

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N$m^{2}$/$C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$, substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/$0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$/0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$, substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/$0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$/0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$, substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/$0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$/0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$/$F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

Answer 2

The addition of vectors and Coulomb's law allows us to find the result for the force on the charge q₃ is:

• The modulus is F = 5.71 N

• The direction is: tes = -65º

Given parameters,

• The value of the charges, q₁= +q, q₂= - 2 q, q₃= -3 q, q₄= -4q

• The value of the charge q=+ 2.4 μC = 2.4 10⁻⁶ C

• The side of the square is d = 33 cm = 0.33 m

To find.

• The force on q3

Electric Force.

The electric force is given by Coulomb's law, which states that the force is proportional to the charges and inversely proportional to the square of the distance.

               F =$k \frac{q_1q_2}{r_{12}^2}$  

where F is the force, k Coulomb's constant, q the charges and r the distance between the charges.

Sum Vectors.

Force is a vector magnitude, so vector algebra must be used for vector addition, an easy and efficient way is to use an analytical method for addition:

• We decompose each vector

• We make the sum of the components

• We construct the resulting vector.

In the attachment we have a diagram of the forces, the sum on each axis is:

x axis

          Fₓ= F₂- F₁ₓ

y axis

          $F_y = -F_4 + F_{1y}$  

Let's use trigonometry to find the component of force.

        cos 45 = $\frac{F_1_x}{F_1}$  

        cis 45 = $\frac{F_1_y}{F_1}$  

        F₁ₓ= F₁ cos 45

        $F_1_y$ = F₁ sin 45

Let's look for the distance between the charges, for charges 2 and 3 charges 4 and 3 the distance of is equal to the side of the square

          r=r₂₃ = r₄₃ = d

The distance between charges 1 and 3 is the diagonal of the square that we can find with the Pythagorean theorem.

       R₁₃² = d² + d²

        R₁₃² = 2d²

We substitute in Coulomb's law.

        F₂₃ = $k \frac{q_2q_3}{d^2}$

        F₄₃ = $k \frac{q_4q_3}{d^2}$

         F₁₃ = $k \frac{q_1q_3}{2d^2 }$  

Let us substitute in the components of the force on each axis.

x axis

         Fₓ= F₂- F₁ₓ

         $F_x = k \frac{q_2q_3}{d^2} - k \frac{q_1q_3}{2d^2} cos 45 \\F_x = k \frac{q_3}{d^2} ( q_2 - \frac{q_1 cos 45}{2})$

y axis

          $F_y = - F_4 + F_1_y \\F_y = k \frac{q_4q_3}{d^2} - k \frac{q_1q_3}{2d^2} sin 45 \\F_y = k \frac{q_3}{d^2} ( -q_4 + \frac{q_1sin 45}{2})$

We substitute in value of each charge.

x axis

           $F_x = k \frac{3q}{d} \ q( 2 - \frac{1 \ cos 45}{2} ) \\F_x = 3k (\frac{q}{d})^2 1.6464$

y axis

         $F_y = k \frac{3q}{d^2} ( -4 + \frac{1 \ sin45}{2} ) \\F_y = 3k (\frac{q}{d})^2 ( -3.64645)$

The resulting vector is

          F = $F_x \hat i + f_y \hat j$  

          F = $3k(\frac{q}{d} )^2 ( 1.6464 \hat i - 3.64645 \hat j )$3k q² /d² ( 1.6464 i^ - 3.64645 j^ )

We calculate.

        F = 3 9 10⁹ ( 2.4 10-6/0.33)² ( 1.6464 i^ - 3.64645 j^ )

         F = 1.428( 1.6464 i⁻ 3.64645 j^

We build the resulting vector, for the module we use the Pythagorean theorem.

          $F = \sqrt{F_x^2 + F_y^2}\\F = 1.428 \ \sqrt{1.6464^2+ 3.64645^2}$

          F = 5.71N

Let's use trigonometry for the angle.

        $tan \theta = \frac{F_y}{F_x}$  

        $\theta = tan^{-1} (\frac{-3.64645}{1.6464} )$  

        θ = -65º

This angle is half clockwise from the positive side of the x-axis.

in conclusion using vector addition and Coulomb's law we can find the result for the force on the charge q3 is:

• The direction is: tes = -65º

• The modulus is F = 5.71 N

Learn more about the electric force here: brainly.com/question/26153267