All categories

2 years ago

Three equal positive charges 'q' are at the corners of an equilateral triangle of side 'a'.

1 answer:

5 0

(a) The location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

(b) The magnitude and direction of the electric field at the top corner due to the two charges at the base is 1.732 kq/a².

<h3>Position where the electric field is zero</h3>

The electric field is zero at the center of the equilateral triangle whose magnitude is equal to √3a/6.

<h3>Electric field at top corner due to two charges at the base</h3>

E = E₁ + E₂

where;

E₁ is electric field at the left base corner
E₂ is electric field at the right base corner

E = kq/a²[(cos 60i + sin 60j) + (-cos 60i + sin 60j)]

E = kq/a²[2(sin 60j)] = 1.732 kq/a²

Thus, the location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

The magnitude and direction of the electric field at the top corner due to the two charges at the base is 1.732 kq/a².

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

You might be interested in

During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1

Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity <u>in radians</u>)

a = $\frac{v^{2}}{r}$

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = $\frac{v^{2}}{r}$

a = $\frac{49^{2}}{0.26}$

a = 9234.6 (m/ $s^{2}$ )

Wow! That's fast!

<u>We now have our answers for a and b:</u>

a. 49 (m/s)

b. 9.2 * $10^{3}$ (m/ $s^{2}$ )

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0

2 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with

ad-work [718]

Answer:

y₀ = 10.625 m

Explanation:

For this exercise we will use the kinematic relations, where the upward direction is positive.

y = y₀ + v₀ t - ½ g t²

in the exercise they indicate the initial velocity v₀ = 8 m / s.

when the rock reaches the ground its height is zero

0 = y₀ + v₀ t - ½ g t²

y₀i = -v₀ t + ½ g t²

let's calculate

y₀ = - 8 2.5 + ½ 9.8 2.5²

y₀ = 10.625 m

7 0

3 years ago

What changes would result in a decrease in the gravitational force between two objects? Check all that apply.

REY [17]

<em>I'm sorry, it says check all that apply, however there are no choices given. You should edit, and add the multiple choice answers.</em>

My Answer:

Well if the masses of two objects were both decreased, it would result in a decrease in the gravitational force. So I guess the two objects masses would need to be decreased.

4 0

3 years ago

In which situation is work not being done?

almond37 [142]

work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)

5 0

3 years ago