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An electron moves in the field produced by a charged sphere of radius 10 cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 × 105 to 2 × 106 m/s. Determine the surface charge (in nC/m2) density of the sphere. (Mass of electron = 9.1 × 10–31 kg)
    Correct answer is between '5.96,5.98'. Can you explain this answer?
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    An electron moves in the field produced by a charged sphere of radius ...
    To solve this problem, we can use the concept of electric potential and potential difference.

    The electric potential at a point due to a charged sphere is given by:
    V = k * (Q / r)

    Where:
    - V is the electric potential at the point
    - k is the Coulomb's constant (9 x 10^9 Nm^2/C^2)
    - Q is the charge of the sphere
    - r is the distance from the center of the sphere to the point

    The potential difference between two points is given by:
    ΔV = V2 - V1

    Where:
    - ΔV is the potential difference
    - V2 is the electric potential at the second point
    - V1 is the electric potential at the first point

    In this problem, the electron moves from a point at a distance of 12 cm from the center of the sphere to a point at a distance of 15 cm from the center of the sphere. Let's calculate the potential difference between these two points.

    First, we need to find the electric potential at each point. Since the sphere is charged, it has a charge Q. The charge Q is not given in the problem, so we cannot calculate the exact potential difference. However, we can calculate the ratio of the potential difference to the initial velocity.

    The electric potential at the first point (r1 = 12 cm) is:
    V1 = k * (Q / r1)

    The electric potential at the second point (r2 = 15 cm) is:
    V2 = k * (Q / r2)

    The potential difference between the two points is:
    ΔV = V2 - V1 = k * (Q / r2) - k * (Q / r1) = k * Q * (1/r2 - 1/r1)

    Given that the initial velocity of the electron is 2 m/s, we have:
    ΔV / initial velocity = (k * Q * (1/r2 - 1/r1)) / initial velocity

    Since k is a constant and Q is the charge of the sphere, we can define a constant C = k * Q. Therefore, we have:
    ΔV / initial velocity = C * (1/r2 - 1/r1)

    So, the potential difference per unit initial velocity is the constant C multiplied by the difference in inverse distances.

    Note: Without knowing the value of Q, we cannot calculate the exact potential difference or the constant C. We can only calculate the ratio of the potential difference to the initial velocity.
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    The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. The transition, so that the de - Broglie wavelength of electron becomes 3 times of its initial value in He+ ion will be

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    An electron moves in the field produced by a charged sphere of radius 10 cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 × 105to 2 × 106m/s. Determine the surface charge (in nC/m2) density of the sphere. (Mass of electron = 9.1 × 10–31kg)Correct answer is between '5.96,5.98'. Can you explain this answer?
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    An electron moves in the field produced by a charged sphere of radius 10 cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 × 105to 2 × 106m/s. Determine the surface charge (in nC/m2) density of the sphere. (Mass of electron = 9.1 × 10–31kg)Correct answer is between '5.96,5.98'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An electron moves in the field produced by a charged sphere of radius 10 cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 × 105to 2 × 106m/s. Determine the surface charge (in nC/m2) density of the sphere. (Mass of electron = 9.1 × 10–31kg)Correct answer is between '5.96,5.98'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron moves in the field produced by a charged sphere of radius 10 cm along the radius connecting the points separated by 12 and 15 cm from the centre of the sphere. The velocity of the electron changes thereby from 2 × 105to 2 × 106m/s. Determine the surface charge (in nC/m2) density of the sphere. (Mass of electron = 9.1 × 10–31kg)Correct answer is between '5.96,5.98'. Can you explain this answer?.
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