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ad-work [718]
3 years ago
9

Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

ectric field at the center of the square? (k=1/4πϵ0=8.99×109N⋅m2/C2 )
Express your answer in terms of the variables d, q, and appropriate constants.
Physics
2 answers:
Rina8888 [55]3 years ago
8 0
<span>this  may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
Vaselesa [24]3 years ago
7 0

Answer:

E = 2kq /d^2

Explanation:

Let d = side of the square

Distance between the center and corner of the square = a

a = √(d^2 + d^2) / 2

a = √2d^2 / 2

a = (√2 * d) /2

a = d /√2

Let the magnitude of an electric field = E

E = kq/a^2

Put a = d/√2

E = kq / (d/√2)^2

E = kq / d^2 / 2

E = 2kq / d^2

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liubo4ka [24]

AnswerNo

However, the acceleration is the opposite of the movement:

Explanation:

8 0
2 years ago
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o-na [289]

Answer:

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Explanation:

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Read 2 more answers
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat
RSB [31]

Answer:

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\

where;

V is the potential difference between the plates

The charge on the plates is given as;

Q = \frac{V\epsilon _0 A}{d}

The energy stored in the capacitor is given as;

E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

3 0
3 years ago
A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and
iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g

The speed at this point would then be

v^2 = a_c r = 2gr = 2*9.8*17 = 333.2

v = \sqrt{333.2} = 18.25 m/s

2. Similarly, if T = mg, then the centripetal acceleration must be

a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0

As the ball has no centripetal acceleration, its speed must also be 0 as well.

6 0
3 years ago
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