1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Deffense [45]
3 years ago
13

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Physics
1 answer:
Bad White [126]3 years ago
6 0

Answer:

\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}

\displaystyle \theta =68^o

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

\displaystyle F=\frac{KQ_1Q_2}{d^2}

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}

\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l

The force F3 is

\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle

F2 is pointed upwards and its horizontal component is zero

\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle

\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

Now we compute the total force

\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}

\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle

\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}

Now we compute the magnitude

\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}

The direction of the total force is given by

\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558

\boxed{\displaystyle \theta =68^o}

You might be interested in
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan
nlexa [21]

Answer:

Explanation:

hi

5 0
2 years ago
Will an object with the density of 0.66g float?
Firdavs [7]
Yes yes it will............
6 0
3 years ago
When four people with a combined mass of 320 kg
ale4655 [162]
Each person is 80kg. 320/4
4 0
3 years ago
A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above
Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + \frac{1}{2}at^2 and plugging in from the y-dimension info:

-10.0=0t+\frac{1}{2}(-9.8)t^2 which simplifies to

-10.0=-4.9t^2 so

t=\sqrt{\frac{-10.0}{-4.9} } which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + \frac{1}{2}at^2 and filling in the x-stuff:

Δx = 7.25(1.4)+\frac{1}{2}(0)(1.4)^2 Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0
3 years ago
Read 2 more answers
Other questions:
  • If a photon has a frequency of 1.15 × 1015 hertz, what is the energy of the photon? Given: Planck's constant is 6.63 × 10-34 jou
    7·2 answers
  • A circular bird feeder of radius R and moment of inertia I is suspended at its center by a thin wire. (The feeder is oriented in
    15·1 answer
  • Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity v0 and glider
    14·1 answer
  • Why does water frozen in the cracks of a rock help to break down the rock? a) Water expands when frozen and chemically forces th
    9·1 answer
  • Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want
    8·2 answers
  • What is it known as when a force is applied to an object for an amount of time?
    11·2 answers
  • Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work
    8·1 answer
  • 2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.
    10·2 answers
  • Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tunin
    12·1 answer
  • All living things inherit similar traits from their
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!