Two Charges and a Test Point
Two point charges sit on the x-axis: q₁ = +6.0 μC at x = 0 and q₂ = −2.0 μC at x = 0.30 m. A third charge q₃ = +1.0 μC is to be placed somewhere on the x-axis. A student computes the electric field at the midpoint x = 0.15 m and finds it points in the +x direction, then argues that the net force on q₃ placed there is therefore toward +x. Separately, the student claims there must be a point between the charges where the net field is exactly zero because the fields 'cancel between opposite charges.' Use k = 9.0×10⁹ N·m²/C².
1. Compute the net electric field at x = 0.15 m, including direction, and evaluate the student's force conclusion for q₃ = +1.0 μC placed there.
Read the worked answer
At x = 0.15 m, distance to q₁ is 0.15 m; to q₂ is 0.15 m. E₁ = k(6.0×10⁻⁶)/(0.15)² = (9.0×10⁹)(6.0×10⁻⁶)/0.0225 = 5.4×10⁴/0.0225 = 2.4×10⁶ N/C, pointing +x (away from positive q₁). E₂: q₂ is negative, so field points toward q₂, i.e., toward +x at this point. Magnitude = (9.0×10⁹)(2.0×10⁻⁶)/0.0225 = 1.8×10⁴/0.0225 = 8.0×10⁵ N/C, also +x. Net E = 3.2×10⁶ N/C in +x. Since q₃ is positive, F = qE points +x. The student's force direction is correct here, but only because q₃ is positive; a negative q₃ would feel force in −x. The reasoning 'field points +x therefore force +x' is only valid for positive charge.
2. The student claims a zero-field point exists between the charges. Determine whether this is true and, if not, state where (if anywhere) the field can vanish on the axis.
Read the worked answer
Between the charges, E₁ points +x and E₂ also points +x (toward the negative charge). They add, never cancel - so no zero point between them. The fields oppose only outside the pair. On the far side beyond q₂ (x > 0.30), the smaller-magnitude charge q₂ is closer, so its field can balance q₁'s: set k(6)/x² = k(2)/(x−0.30)² → 6(x−0.30)² = 2x² → 3(x−0.30)² = x² → √3(x−0.30) = x → x(√3−1) = 0.30√3 → x ≈ 0.519/0.732 ≈ 0.71 m. On the far left (x < 0), q₁ is closer and larger, so it always dominates; no zero there. The zero-field point is at x ≈ 0.71 m.
3. Without recomputing from scratch, describe how the electric potential (not field) behaves at x = 0.15 m relative to infinity, and whether a zero-potential point exists between the charges.
Read the worked answer
Potential is a scalar: V = kq₁/r₁ + kq₂/r₂. At x = 0.15 m, r₁ = r₂ = 0.15 m, so V = (9.0×10⁹)(6.0×10⁻⁶ − 2.0×10⁻⁶)/0.15 = (9.0×10⁹)(4.0×10⁻⁶)/0.15 ≈ 2.4×10⁵ V, positive. Between the charges, V = k(6/x − 2/(0.30−x)); as x→0⁺ it → +∞, as x→0.30⁻ it → −∞, so by continuity V passes through zero somewhere between - unlike the field, which never cancels there. Setting 6/x = 2/(0.30−x) gives 6(0.30−x) = 2x → 1.8 = 8x → x = 0.225 m. The contrast (zero potential but nonzero field) is the key conceptual point.