Two point charges, 4.0 µC and -2.0 µC, are placed 5.2 cm apart on the x axis, su

Question

Two point charges, 4.0 µC and -2.0 µC, are placed 5.2 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 4.0 µC charge is at x = 5.2 cm. (a) At what point(s) along the x axis is the electric field zero? (If there is no point where E = 0 in a region, enter "0" in that box.)

(b) At what point(s) along the x axis is the potential zero? Let V = 0 at r = . (If there is no point where V = 0 in a region, enter "0" in that box.)

Explanation / Answer

(a)

Because the charges have opposite sign, the zero-field point can't be between them (where the field always points to the negative charge). Also, the zero-field point can't be closer to the charge with the larger magnitude, or it will always dominate. For this problem, then, the zero-field point is on the negative x axis.

Therefore "x < 0" is the correct choice.


If its distance from the origin is "x," then we need
k*2µC / x² = k*4µc / (5.2 + x)² for x in cm. k*1µ cancels, leaving
2 / x² = 4 / (5.2+x)²
2(5.2+x)² = 4x²
2(27.04 + 10.4x + x²) = 4x²
x² - 10.4x - 27.04 = 0
is quadratic with roots at
x = -2.15 cm not possible; the way I've defined x it must be > 0
x = 12.55 cm to the left of the origin

(b) k*2µC / x = k*4µC / (5.2 + x)
2(5.2+x) = 4x
5.2 + x - 2x = 0
x = 5.2cm or -5.2cm
The way I set this up, it means 5.2 cm to the RIGHT of the origin, so
0 < x < 5.2 cm looks right

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.