1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Field concept

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello. Im having trouble understanding this particular statement found in one of my textbooks.

    " Work done to move a charge along a line perpendicular to the field is zero"

    3. The attempt at a solution

    Field is a space around a charge where if we place any other charge, will result in it experiencing a force.

    Field is a 3d space i.e., all the point inside it is able to exert a force on an external charge. And the field lines at a point are just a representation of the force a positive(can it be negative too?) charge experience when placed at that point.

    Only an external electric(or due to a magnetic field but that is due to the electric field in turn produced by the magnetic field) field can move a charge inside another electric field. Inorder to do that it must work against the force that keeps the charge in the field. This work changes to potential energy of the charge that got separated from the field.

    Now, if the work done is zero that implies there was no electric field/force of attraction which must prevent this motion along that path of motion. It is true that there is no component of a vector along a line perpendicular to a vector(field line). But wouldnt there be other field lines (or their components) in this path i.e., perpendicular to a particular field line, since the field is 3d concept? Or are all field lines aligned in a particular direction?

    Kindly correct me if any of the above statements is wrong. Thanks in advance!
  2. jcsd
  3. Jan 7, 2016 #2


    User Avatar
    Science Advisor

    The field lines are indeed aligned. The field value at any particular point is well defined and only points in a single direction with a single magnitude. The direction and magnitude of the field values at nearby points are not very different from this. In other words, the fields that one encounters are normally continuous.

    In mathematics, a "vector field" amounts to a function that assigns a single vector value to each and every point in a region. The work done along a path through a vector field is the sum (actually a "path integral") of the dot product of the field value at each tiny incremental segment along that path multiplied by the incremental length of that segment. If the field is not continuous then such an integral will not exist.

    If a path is at right angles to the field direction at every point along its length then the contribution of each incremental segment is zero and and the sum (or integral) will also be zero.
  4. Jan 7, 2016 #3


    User Avatar
    Homework Helper

    See if this helps. Consider a charge +q, emitting field lines in all the directions(3D, as you say). If you take any point at a distance r from the charge in any direction, potential at that point will be kq/r. This means, if you draw a hypothetical sphere with radius r with q at it's center, all the points on the surface of the sphere will be at the same potential. Hence there won't be any field between any two points on the surface.The field lines are normal to the surface, pointing radially outwards. As r increases, strength of this field goes on reducing in the same direction i.e.potential goes on decreasing. Now, if you want to move another +q1 charge from r to the center charge along the radius of the sphere, q1will feel a very strong opposing force(hence, field) but if the same q1 were to be moved anywhere on the surface of the sphere, it wouldn't experience any opposing field along it's way since all the points there are at the
    same potential. This means component of electric field normal to it's direction(i.e. tangential to the sphere) is 0.
  5. Jan 8, 2016 #4


    User Avatar

    Staff: Mentor

    To move a charge along any line of equipotential takes zero effort. Only if you move charge between points of differing potentials is work done.
  6. Jan 8, 2016 #5
    Thank you guys.

    Isnt work = force * distance? By field value did you mean force? Thanks for giving insights into field concept.

    Thanks a lot. You have actually cleared my doubt on this particular issue. But in normal cases charges wont exist alone....there will be a group of them no? So an equipotential surface(surface with equal r from charge source) with respect to one will not be due to other charge's effect as charges exist at different positions in space? So does equipotential surface exist, practically..if so are there any other instance other than a single charge scenario?
    Thanks. Iwas having trouble getting concept of equipotential surface. But previous reply solved it i think...
  7. Jan 8, 2016 #6


    User Avatar
    Homework Helper

    Putting an isolated charge at a single point in space is practically very difficult, true. In practice, however, equipotentials do exist. If you put some charge on a conductor, it spreads uniformly on it's surface. Hence, all the points on and inside the conductor are at the same potential, making electric field inside the conductor 0.
  8. Jan 8, 2016 #7


    User Avatar
    Science Advisor

    Yes, work is force * distance. But it is force times parallel distance. If you push a hockey puck across the ice horizontally, you are doing zero work against the vertical force of gravity. If you lift a box vertically you are doing work against the vertical force of gravity. If you push a box up an inclined slope, then the work you are doing depends on the angle of the slope.

    The "vector dot product" captures this effect. The dot product of two vectors is given by the product of their magnitudes multiplied by the cosine of the angle between them. If the vectors are at right angles then that cosine is zero. That is why the "path integral" taken through a vector field uses the dot product -- to capture this effect.

    Yes, in order for the path integral through a vector field to yield a result that is "work", you need for the field values to have units of force. In the case at hand, we have an electrical field in units of force per unit charge. If you were to compute the integral over a path you would wind up with "work per unit charge" rather than just "work". Work per unit charge is measured in units of Volts -- so the path integral yields a voltage rather than a quantity of energy. (Well spotted!)
  9. Jan 8, 2016 #8


    User Avatar
    Homework Helper

    I believe yes, it's the force. If it were magnitude of electric field then the integral would give the value of potential difference between the two points. Electrostatic field is conservative, just like gravitational field. No matter what path a charge takes to go from A to B, it will end up with the same electrostatic potential energy and experience the same potential difference.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Field concept
  1. Field concept (Replies: 9)