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Electric Potential and Electric Field

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Which point in the electric field in the diagram is at the lowest potential?

    Figure: http://i131.photobucket.com/albums/p289/SoaringCrane/Tip24_fig3.gif

    a.Point 1

    b.Point 2

    c.Point 3

    d.Point 4

    e.Point 5

    2. Relevant equations

    See below.

    3. The attempt at a solution

    Equipotential lines are perpendicular to electric field lines, so dots in the same column have the same potential. Electric field lines go from high voltage to low voltage, so dots at the near right have the lowest potential? Rightmost is Point 3?

    1. The problem statement, all variables and given/known data

    Charges Q and q (Q is not equal to q), separated by a distance d, produce a potential V_p at point P.

    Figure: http://i131.photobucket.com/albums/p289/SoaringCrane/Tip24_fig1.gif

    This means that

    a.no force is acting on a test charge placed at point P.

    b.Q and q must have the same sign.

    c.the electric field must be zero at point P.

    d.the net work in bringing Q to a distance d from q is zero.

    e.the net work needed to bring a charge from infinity to point P is zero.

    2. Relevant equations

    See below.

    3. The attempt at a solution

    Following the fact that V/m = N/C, the electric field will also be 0???

    1. The problem statement, all variables and given/known data

    Charges +Q and -Q are arranged at the corners of a square as shown.

    Figure: http://i131.photobucket.com/albums/p289/SoaringCrane/Tip24_fig2.gif

    When the electric field and the electric potential V are determined at P, the center of the square, we find that

    a. E no equal 0; V>0
    b. E = 0; V = 0
    c. E = 0; V>0
    d. E no equal 0; V<0

    2. Relevant equations

    See below.

    3. The attempt at a solution

    Well, if the vectors are drawn from the center, they cancel out so E = 0. If E=0, then wouldn’t the potential also be 0?

    Last edited: Feb 5, 2007
  2. jcsd
  3. Feb 5, 2007 #2
    Well, I'm studying for an exam over this material myself...I think your first problem is correct...but I'm not sure about #2 or 3. For number 2, if you select C, then doesnt that make a also correct? In number 3, you field can be 0, but it does not mean that the potential is 0.

    Like I said, I'm studying for this myself so this is great practice!
  4. Feb 5, 2007 #3

    Please be patient enough the read the full text as it took me a while to type them for you!

    Question 1

    The key idea behind Electric Potential (or any potential for that sake) is the conservative nature of the field; in this case the electric field.

    Conservative fields are those in which any work done against the direction of field is stored as potential of the body. In case of electric fields, the field tends to push a positively charged particle in the direction of field. Thus a charge kept in a field experiences a force. When the particle is displaced in a direction opposite to that of this force, work is done.

    this work, W =F.S=-FS (coz S is opposite to F)

    If this process takes place in such a manner that no net acceleration takes place at any instant, then total work done to displace the particle ,which is integral -F.dS (because Force varies with position), is stored as the potential energy of the particle. Deriving, we would get V proportional to 1/r and hence greater the value of r, lesser the potential.

    In theory this actually means that least work need to be done to bring a test charge to point 3 than the other points.

    Hope its Clear!
  5. Feb 5, 2007 #4
    I don't think I quite understand Question 2. The statement 'electric field is zero' can only mean that force experienced by a charge placed at that point is zero. There would be a potential at any point between 2 unequal charges whether or not the net field there is zero. So I would like to suggest that, for the given data, your answer is incorrect.

    For question 3, E can be determined by resolving the E vectors due to different charges. However it is incorrect to conclude here again that V=0, simply because E=0. Instead you need only add the potential due to each charge directly because potential is a scalar. Thus we can immediately conclude that V=0, since 2 charges are -ve and the other 2 are positive and all are equidistant from the center.
  6. Feb 5, 2007 #5
    Are u suggesting that the same potential V_p is produced by both the charges or V_p is the resultant potential?

    If its the first case the E would indeed be zero at that point!
  7. Feb 5, 2007 #6
    Now that you introduced this question, I think that the problem addresses V_p as the latter, the resultant potential? It is supposed to be V_p = 0.

    Charges Q and q (Q is not equal to q), separated by a distance d, produce a potential V_p = 0 at point P.
    Last edited: Feb 5, 2007
  8. Feb 5, 2007 #7
    I'm sorry, but I typed the middle question haphazardly. The statement in bold is the correct part. I have eliminated the first three choices (hopefully, I did this right??) -

    For a. and c., if there is 0 electric field intensity, the force will also be 0.

    For b., the charges have to have opposite signs since the potential difference is not a vector sum.

    If the above reasonings are not completely bogus, then a 0 potential indicates the work that must done to move a charge from some position to point P?
  9. Feb 6, 2007 #8
    Your reasoning c0ncerning the sign of Q and q is correct. But in that case electric field will never be zero because The resultant field vector will always be directed from +ve to-ve charge and field cannot be zero at ANY point between Q and q!

    Thus, based on your corrected statement, the answer would be (e).
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