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Dafna1 [17]
3 years ago
15

An electric field of magnitude e is measured at a distance r from a point charge q. if the charge is doubled to 2q, and the elec

tric field is now measured at a distance of 2r from the charge, what is the value of the new electric field?
Physics
1 answer:
pav-90 [236]3 years ago
5 0
Electric field is proportional to the charge and inversely proportional to the square of distance:
E\propto\frac{Q}{r^{2}}
With charge 2q and distance 2r, the electric field is proportional to:
E_{q}\propto\frac{2q}{(2r)^{2}}=\frac{2q}{4r^{2}}=\frac{1}{2}\frac{q}{r^{2}}
The new electric field is half of that of the original measured field.
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Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.
olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

     E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

   E1 = k (-q) / a²

   E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

    E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

    Et = E1 + E2

    Et = kq / a² + kq / a²

    Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

     E1 = k q / (R + a)²

     E2 = kq / (R-a)²

     Et = kq [1 / (R + a)² + 1 / (R-a)²]

     Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

     Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that  R> a we can despise in the patents "a"

     (R² + a²) = R² (1+ a² / R²) ≈ R²

     (R + a)² = R² (1 + a / R)² ≈ R²

     (R- a)²  = R² (1-a / R)² ≈ R²

Substituting in the total electric field

     Et = kq {2 R²) / [R²R²]}

     Et =kq 2 / R²

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vivado [14]

Answer: F = ma,

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Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV
forsale [732]

Answer:

(a) \lambda=1.227\ A

(b) \lambda=0.388\ A

(c) \lambda=0.038\ A

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

\lambda=\dfrac{h}{\sqrt{2meV} }

Where

m and e are the mass of and charge on an electron

On solving,

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

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\lambda=\dfrac{12.27}{\sqrt{100} }\ A

\lambda=1.227\ A

(b) V = 1 kV = 1000 V

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

\lambda=\dfrac{12.27}{\sqrt{1000} }\ A

\lambda=0.388\ A

(c) If V=100\ kV=10^5\ V

\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A

\lambda=0.038\ A

Hence, this is the required solution.

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