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bulgar [2K]
2 years ago
11

What is the magnitude of the electric field at the dot in the figure? (figure 1)

Physics
1 answer:
blondinia [14]2 years ago
5 0

The magnitude of the electric field at the dot is : 10⁴ v/m

<h3>What is electric field?</h3>

Electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field

Determine the magnitude of the electric field at the dot

change in voltage  = E .d

100 - 0 = E \times ( 1 \times 10^{-2}\  m ) ----- ( 1 )

From equation ( 1 )

The magnitude of electric field will be given as

E = \dfrac{100 }{ ( 1 \times  10^{-2})\ }

E = 10^4\ \ \frac {v}{   m}

Hence we can conclude that The magnitude of the electric field at the dot is :  10⁴ v/m

Learn more about electric field :

brainly.com/question/14372859

#SPJ4

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If an element has an atomic number of 28, how many protons does it have in its nucleus?
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I think the answer is 28 protons
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3 years ago
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The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra
Norma-Jean [14]

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

4 0
3 years ago
HELP THIS LAST DAY!!!
Alex_Xolod [135]

Answer:

The proton has much greater mass

Explanation:

  • Protons and electrons are part of an atom
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8 0
2 years ago
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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

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Answer:   Work W = 0

Explanation: Work W = F·s. Because rock does not move, s = 0 and

work done is zero.

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