Suppose we put two charged beads (point charges) on a taut string, some distance
ID: 2078030 • Letter: S
Question
Suppose we put two charged beads (point charges) on a taut string, some distance apart, but fixed in place (immovable). One bead's charge is positive, the other bead's charge is negative, and the charges' magnitudes are unequal. No other charges are nearby these two, and we'll let the string serve as the x axis.
Now choose some position A on the string, other than the positions of the beads. Position A should not be the midpoint between the beads, but any other position on the same axis is fine.
Question: if you reversed only the sign of one of the two bead's charges, would the net electric field at A increase, decrease, or remain the same? What about the net electric potential?
Note: if you find yourself calculating forces (with Coulomb's law), you're headed in the wrong direction. Same goes for calculating electric potential energies. This question concerns electric fields (E) and electric potential (V) only, and no forces or potential energies whatsoever are involved in the solution, so none should be included.
To answer this question by calculation, start by writing a Setup describing an experiment matching the above description. Then solve it, using only equations from the PHYS 1201 Equation Sheet. Write up the complete solution and submit it as a single document in PDF, DOC, or DOCX format using the CX1 link above.
Include two sketches in the Experiment section, each showing the x axis, the two point charges, position A, and the directions of the electric fields at A due to each charge. One sketch is for the original case, and the other sketch shows the situation when the sign of one of the charges is reversed. Label all parts of the sketches clearly. Please read all assignment information in this folder before starting to work, particularly the CX example and the grading criteria.
EQUATIONS
Metric prefixes
f =1015 p=1012 n =109 =106 m=103 k =103 M =106 G =109
Electric charge and field
E = ke q / r 2 F= q E F = ke q1q2 / r2 ke =8.99×109 Nm 2 /C 2 e =1.60×1019 C me =9.11×1031 kg mP =1.673×1027 kg
Explanation / Answer
When we reverse sign of one of the charges, then they both become of same sign.
As the point A is in between the two beads now the electic field at point A due to two beads are in opposite direction (initially in the direction) so its magnitude will decrease.
For potential it is just a alzebaic sum so its magnitude will increase as we are going to add both of them.