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    (Original post by PhysicsIP2016)
    As there is a potential difference, there is now a uniform electric field which acts across the wings of the plane from right to left. This exerts a force on the charged particles so now the positive charges are attracted towards the left hand side of the plane where the magnetic field force initially moved them to the right hand wing
    Ah, I see! Thank you so much
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    Hi. Need some help here.

    https://ad57adc4f6eb5ea42b541057f162...%20Physics.pdf

    6- Why is the answer B instead of A?
    13(cii)-In such questions, do we mention about centripetal force instead of contact?
    14c. If the charge is increased in one of the balls(with both having the same sign), so shouldn't there be a greater repulsion between them?
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    (Original post by sabahshahed294)
    Hi. Need some help here.

    https://ad57adc4f6eb5ea42b541057f162...%20Physics.pdf

    6- Why is the answer B instead of A?
    13(cii)-In such questions, do we mention about centripetal force instead of contact?
    14c. If the charge is increased in one of the balls(with both having the same sign), so shouldn't there be a greater repulsion between them?
    For 6, the spheres have identical momentum but in the opposite direction. So total momentum is mv - mv = 0.
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    (Original post by Mowerharvey)
    For 6, the spheres have identical momentum but in the opposite direction. So total momentum is mv - mv = 0.
    Cool! Thanks
    Ad wb the rest of them? any idea?
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    (Original post by sabahshahed294)
    Cool! Thanks
    Ad wb the rest of them? any idea?
    For questions like 13(bii) I'm sure you need to mention contact force and centripetal force.

    14(c) yes if the charge of one sphere increases then the force of repulsion acting on both spheres will increase by the same amount, and the angle will increase as well
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    (Original post by sabahshahed294)
    6- Why is the answer B instead of A?
    13(cii)-In such questions, do we mention about centripetal force instead of contact?
    14c. If the charge is increased in one of the balls(with both having the same sign), so shouldn't there be a greater repulsion between them?
    For 13cii):
    The centripetal force is the resultant force of the weight + contact (reaction) force, which causes the cabin to accelerate. It thus always acts towards the centre of the circle.
    At A: Centripetal force = mg - r
    The weight must be greater than the reaction force in order to provide a centripetal force, hence the reaction force is at a minimun.
    At C: Centripetal force = r - mg
    Here, the reaction force must be greater than the weight, hence the reaction force is at a maximum.

    For 14c):
    Yes, there is a greater repulsion between them, but an equal force acts on both balls. Therefore, they would be suspended at a greater angle to the vertical, but this angle would be the same for both balls.
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    Thank you both!
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    Hello, Can anyone help out in this paper?

    http://qualifications.pearson.com/co...e_20130613.pdf

    5 and 10-Didn't get the logic behind these two questions.
    16(c)- How do you make the equation here for tan theta?
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    (Original post by sabahshahed294)
    Hello, Can anyone help out in this paper?

    http://qualifications.pearson.com/co...e_20130613.pdf

    5 and 10-Didn't get the logic behind these two questions.
    16(c)- How do you make the equation here for tan theta?
    For 5), the switch is opened and the capacitor disharges through the resistor exponentially, hence the exponential curve at the beginning. When the switch is closed, the capacitor will charge instantaneously. This happens because the capacitor is not charging through the resistor, which makes T = RC = 0 therefore the time it takes to charge the capacitor is 0. So the line at the point when the switch is closed is vertical (answer b)

    For 10) I would like to know the logic behind this question too
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    (Original post by Mowerharvey)
    For 5), the switch is opened and the capacitor disharges through the resistor exponentially, hence the exponential curve at the beginning. When the switch is closed, the capacitor will charge instantaneously. This happens because the capacitor is not charging through the resistor, which makes T = RC = 0 therefore the time it charges to charge the capacitor is 0. So the line at the point when the switch is closed is vertical (answer b)

    For 10) I would like to know the logic behind this question too
    Thanks for your help in Q5.
    What about Q 16(c)? Any idea how to make the equation?
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    (Original post by sabahshahed294)
    Thanks for your help in Q5.
    What about Q 16(c)? Any idea how to make the equation?
    I'm stuck on 10 too - I would have thought that a higher speed = smaller wavelength = more diffraction. Unless perhaps maximum diffraction would create a clear image? Afraid I'm lost on that one too!

    For 16c -
    Sin theta / cos theta = tan theta
    So dividing equation for vertical motion (Nsin theta = mg) by horizontal motion (Ncos theta = mv^2/r) gives tan theta = gr/v^2
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    (Original post by candycake)
    I'm stuck on 10 too - I would have thought that a higher speed = smaller wavelength = more diffraction. Unless perhaps maximum diffraction would create a clear image? Afraid I'm lost on that one too!

    For 16c -
    Sin theta / cos theta = tan theta
    So dividing equation for vertical motion (Nsin theta = mg) by horizontal motion (Ncos theta = mv^2/r) gives tan theta = gr/v^2
    No issues!
    And thank you for your help!
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    (Original post by candycake)
    I'm stuck on 10 too - I would have thought that a higher speed = smaller wavelength = more diffraction. Unless perhaps maximum diffraction would create a clear image? Afraid I'm lost on that one too!

    For 16c -
    Sin theta / cos theta = tan theta
    So dividing equation for vertical motion (Nsin theta = mg) by horizontal motion (Ncos theta = mv^2/r) gives tan theta = gr/v^2
    Isn't the smaller wavelength needed for less diffraction? Diffraction only occurs at a maximum if the gap is equal to the wavelength and at a minimum if the gap is greater than the size of the wavelength of the wave, so by having a smaller wavelength isn't diffraction minimised?
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    (Original post by PhysicsIP2016)
    Isn't the smaller wavelength needed for less diffraction? Diffraction only occurs at a maximum if the gap is equal to the wavelength and at a minimum if the gap is greater than the size of the wavelength of the wave, so by having a smaller wavelength isn't diffraction minimised?
    Been wondering about that but not getting the link behind it with this question!
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    (Original post by PhysicsIP2016)
    Isn't the smaller wavelength needed for less diffraction? Diffraction only occurs at a maximum if the gap is equal to the wavelength and at a minimum if the gap is greater than the size of the wavelength of the wave, so by having a smaller wavelength isn't diffraction minimised?
    The question asks how "to reduce this effect when viewing a smaller object". If the object is smaller as well as the electron wavelength, does this not worsen the diffraction? I have probably misinterpreted/misunderstood the question though!
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    (Original post by candycake)
    The question asks how "to reduce this effect when viewing a smaller object". If the object is smaller as well as the electron wavelength, does this not worsen the diffraction? I have probably misinterpreted/misunderstood the question though!
    Oh I see the confusion now! You're right that there would be more diffraction in smaller objects compared to larger objects, but if you were to increase the size of the wavelength then the electron beam would not pass through the object at all.
    To compensate for making the object smaller, the associated de Broglie wavelength must be smaller than the size of the 'gap' to reduce the effect of any diffraction.
    sabahshahed294 does this help?
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    (Original post by PhysicsIP2016)
    Oh I see the confusion now! You're right that there would be more diffraction in smaller objects compared to larger objects, but if you were to increase the size of the wavelength then the electron beam would not pass through the object at all.
    To compensate for making the object smaller, the associated de Broglie wavelength must be smaller than the size of the 'gap' to reduce the effect of any diffraction.
    sabahshahed294 does this help?
    Thank you!
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    (Original post by PhysicsIP2016)
    Oh I see the confusion now! You're right that there would be more diffraction in smaller objects compared to larger objects, but if you were to increase the size of the wavelength then the electron beam would not pass through the object at all.
    To compensate for making the object smaller, the associated de Broglie wavelength must be smaller than the size of the 'gap' to reduce the effect of any diffraction.
    sabahshahed294 does this help?
    Yes, it does. Thank you!
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    (Original post by PhysicsIP2016)
    Oh I see the confusion now! You're right that there would be more diffraction in smaller objects compared to larger objects, but if you were to increase the size of the wavelength then the electron beam would not pass through the object at all.
    To compensate for making the object smaller, the associated de Broglie wavelength must be smaller than the size of the 'gap' to reduce the effect of any diffraction.
    sabahshahed294 does this help?
    So this means the electrons need to move faster so that the electrons associated wavelength is smaller?
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    (Original post by Mowerharvey)
    So this means the electrons need to move faster so that the electrons associated wavelength is smaller?
    yes
 
 
 
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