20170314

Physics quiz question: total electric field direction between two charges

Physics 205B Quiz 3, spring semester 2017
Cuesta College, San Luis Obispo, CA

Two point charges are held at fixed positions. A +4.0 µC charge is at x = –1.0 cm, and a +8.0 µC charge is at x = +2.0 cm. The electric field at x = 0 is directed to the:
(A) left.
(B) right.
(C) (There is no direction, as the electric field would be zero.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (B)

The electric field vector at x = 0 due to the +4.0 µC source charge points to the right, out away from this positive source charge. The magnitude of the electric field at x = 0 due to the +4.0 µC source charge is:

E4 = k·|Q|/r2,

E4 = k·|+4.0 µC|/(1.0 cm)2,

E4 = (4.0 µC/cm2k.

The electric field vector at x = 0 due to the +8.0 µC source charge points to the left, out away from this positive source charge. The magnitude of the electric field at at x = 0 due to the +8.0 µC source charge is:

E8 = k·|Q|/r2,

E8 = k·|+8.0 µC|/(2.0 cm)2,

E8 = (2.0 µC/cm2k.

The total electric field at x = 0 due to both source charges is the vector addition of the individual electric field vectors at that location. Since the electric fields due to the +4.0 µC source charge and the +8.0 µC source charge point in opposite directions at x = 0, the direction of the total electric field there would point to the right (as the magnitude of E4, which points to the right is greater than the magnitude of E8, which points to the left).

(As an aside, the magnitude of the total electric field at x = 0 would be the difference of the two individual electric field magnitudes, as the electric fields due to the +4.0 µC source charge and the +8.0 µC source charge point in opposite directions:

E = |E8E4|,

E = (4.0 µC/cm2k – (2.0 µC/cm2k,

E = (2.0 µC/cm2k,

E = (2.0 µC/cm2)((1 C)/(106 µC))·((100 cm)/(1 m))2·(8.99×109 N·m2/C2),

E = 179,800,000 N/C,

or to two significant figures, the magnitude of the total electrical field at x = 0 is 1.8×108 N/C.)

Sections 30882, 30883
Exam code: quiz03d3St
(A) : 10 students
(B) : 13 students
(C) : 5 students
(D) : 0 students

Success level: 46%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.40

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