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goldenfox [79]
2 years ago
10

Two long straight parallel lines of charge, #1 and #2, carry positive charge per unit lengths of λ1 and λ2, respectively. λ1 &gt

; λ2. The locus of points where the electric field is zero in this case is
Physics
1 answer:
Stella [2.4K]2 years ago
3 0

Answer:

Somewhere between the two wires, but closer to the wire carrying λ₂

Explanation:

Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².

Electric Fied due to an electric charge is a vector and its direction is  such that  if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)

According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires  are opposite

In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.

As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.

But for points closer to wire with λ₂  ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance  to get equals E and then Ef = 0

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Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le
vlabodo [156]

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

8 0
1 year ago
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe
My name is Ann [436]
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
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6 0
3 years ago
How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or
Ostrovityanka [42]

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

r' = \frac{r}{2}

So the new force will be

F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F

So, the force will quadruple.

4 0
3 years ago
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Anastaziya [24]
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