Anonymous asked in Science & MathematicsPhysics · 6 months ago

More physics, please help me ... (Preparation for an exam.)?

This topic is enormously hard for me and in the main text book which I use, all the provided material is introduced literally in the most difficult way. :(((

May somebody help me to figure out...

3. In the exam I will probably me asked to the explain the diagrams and graphics, so I have found this task the point of which is to show the direction of the electric field at the point A and then at the point B but how can I find out that if there is no information if the particle (A or B) is positively charged or negatively….


4. Lets assume that the Earth and the Moon have received the same charge (+q), How large should the charge be so that the gravitational force between the Earth and the Moon is equivalent to the “electric removal force”?

This task I don’t understand at all. I have tried to count everything by different formulas and by using the masses which the google gave me but the results are very weird and cannot be right for sure. But I guess it is quit the same as the second question in theoretical part and I did not understand it, in the first place..... as I have mentioned before. T…T

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2 Answers

  • 6 months ago
    Favorite Answer

    "direction of the electric field at the point A and then at the point B"

    look at the definition of electric field. It has nothing to do with any charges at A or B. You insert an test charge at the point and measure the force.


    distance to + charge is X² = d² + s²

    assume each charge is Q and they are the same magnitude

    E = kQ/X² and direction is to the left pointed slightly down

    distance to – charge is X² = d² + s²

    E = kQ/X² and direction is to the right pointed slightly down

    see below. The two vectors add up to one smaller one pointing down.


    similar, the two forces are colinear, but the one pointing up is slightly larger as the – charge is closest. Net is a small vector pointing up.

    Electric field

    The strength or magnitude of the field at a given point

    is defined as the force that would be exerted on a

    positive test charge of 1 coulomb placed at that point;

    the direction of the field is given by the direction of

    that force.

    E = F/Q = kQ/r²

    Q = F/E F = QE

    in Newtons/coulomb OR volts/meter

    k = 1/4πε₀ = 8.99e9 Nm²/C²

    re the other problem, just use:

    Coulomb's law, force of attraction/repulsion

    F = kQ₁Q₂/r²

    Q₁ and Q₂ are the charges in coulombs

    F is force in newtons

    r is separation in meters

    k = 8.99e9 Nm²/C²

    k = 1/(4πε₀)

    Gravitational attraction in newtons

    F = G m₁m₂/r²

    G = 6.674e-11 m³/kgs²

    m₁ and m₂ are the masses of the two objects in kg

    r is the distance in meters between their centers

    kQ²/r² = G m₁m₂/r²

    Q² = G m₁m₂/k

    plug in the numbers

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  • 6 months ago


    Is it or is it not *point* A? Why would there suddenly be a particle at A and why would it matter?

    At A

    If the red and blue charge have the same magnitude (don't know if you left that our or what) then obviously the horizontal components of the field caused by them will cancel out while the vertical components both point down (field caused by + charge points radially away from it while for - charge it is towards it)

    So resultant field will be vertically down

    At B

    The field caused by - charge has a greater magnitude than by +.

    Since the larger field points up, it will cancel with the smaller downward-pointing for a resultant upwards field.


    Yeah no idea what that means

    Just try GMm / r^2 = kqq / r^2

    q = ± sqrt(GMm/k)

    M and m are masses of Earth and Moon

    In this situation the Moon and Earth could just stay at the same distance from each other without orbiting for a short time, and would soon get separated by Sun's gravity.

    • D.Filkin6 months agoReport

      Nope, there were no more information - only this task "show direction" and the picture. :(
      But still thank you! I

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