- #1
casanova2528
- 52
- 0
there are 2 situations that need explanation. first, the general formula for electric potential as you take potential = 0 at infinity as a reference.
Second, the general formula for capacitance on a parallel plate.
In situation one, the negative sign does NOT DISAPPEAR, and in the second situation...the negative sign DISAPPEARS! I'm not too sure why.
in the first situation...
the electric field from a positive charge emanates outward toward infinity. The work of the electric field created by the positive charge is NEGATIVE as you bring a positive test charge from infinity to the positive charge, and V = - integral of E vector dot dL vector. The negative sign tells us that the work of the positive electric field from the positive source charge is negative and the
-W = U.
However, as you go from infinity to the positive charge, the dot product between E and dl creates a cos 180 degrees(the dL vector is in the opposite direction of the E vector from the positive source charge). Thus, shouldn't V = + integral of magnitude of E vector times dL vector? then, for a point charge... V = - (k)(q)/r as you take infinity as a reference point. BUT THIS EQUATION IN NOT CONSISTENT WITH THE IDEA THAT IT CREATES POSITIVE ELECTRIC POTENTIAL AS YOU BRING A TEST CHARGE CLOSER TO THE SOURCE POINT CHARGE. What am I missing?in the second situation...
the electric field between a parallel plate (one with +q and the other with -q) from Gauss's law is (q) (permittivity) / Area
V = - integral of E vector dot dL vector to figure out the electric potential created when taking a positive test charge from negative plate to positive plate. This makes sense because the negative sign indicates the work done by the positive electric field as one brings the positive test charge from the negative plate to the positive plate.
However, the negative sign disappears from taking the dot product of the general electric potential formula.
this gives us...
V = Ed what gives?! why can we not do this in situation 1?
Second, the general formula for capacitance on a parallel plate.
In situation one, the negative sign does NOT DISAPPEAR, and in the second situation...the negative sign DISAPPEARS! I'm not too sure why.
in the first situation...
the electric field from a positive charge emanates outward toward infinity. The work of the electric field created by the positive charge is NEGATIVE as you bring a positive test charge from infinity to the positive charge, and V = - integral of E vector dot dL vector. The negative sign tells us that the work of the positive electric field from the positive source charge is negative and the
-W = U.
However, as you go from infinity to the positive charge, the dot product between E and dl creates a cos 180 degrees(the dL vector is in the opposite direction of the E vector from the positive source charge). Thus, shouldn't V = + integral of magnitude of E vector times dL vector? then, for a point charge... V = - (k)(q)/r as you take infinity as a reference point. BUT THIS EQUATION IN NOT CONSISTENT WITH THE IDEA THAT IT CREATES POSITIVE ELECTRIC POTENTIAL AS YOU BRING A TEST CHARGE CLOSER TO THE SOURCE POINT CHARGE. What am I missing?in the second situation...
the electric field between a parallel plate (one with +q and the other with -q) from Gauss's law is (q) (permittivity) / Area
V = - integral of E vector dot dL vector to figure out the electric potential created when taking a positive test charge from negative plate to positive plate. This makes sense because the negative sign indicates the work done by the positive electric field as one brings the positive test charge from the negative plate to the positive plate.
However, the negative sign disappears from taking the dot product of the general electric potential formula.
this gives us...
V = Ed what gives?! why can we not do this in situation 1?