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2 years ago

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.

Physics

2 answers:

kobusy [5.1K]2 years ago

8 0

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

Colt1911 [192]2 years ago

5 0

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

<h3>Electric Force. </h3>

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.

<h3>Sum Vectors. </h3>

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

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

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1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )