Examkin

Question

Difficulty: EasyElectrostatics and Electric Charges

Two equal positive point charges, each of magnitude 2.0×10−6 C2.0 \times 10^{-6}\text{ C}, are placed 0.2 m0.2\text{ m} apart in a vacuum. What is the magnitude of the net electric field intensity at the midpoint between the two charges? (Take Coulomb's constant k=9.0×109 N m2 C−2k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2})

  1. 0 N C−10\text{ N C}^{-1}Answer
  2. B
    1.8×106 N C−11.8 \times 10^6\text{ N C}^{-1}
  3. C
    3.6×106 N C−13.6 \times 10^6\text{ N C}^{-1}
  4. D
    7.2×106 N C−17.2 \times 10^6\text{ N C}^{-1}

Answer

The net electric field intensity at the midpoint is 0 N C−10\text{ N C}^{-1}.
At the midpoint between two identical positive charges, the electric field created by each charge has the exact same magnitude because the charges and distances are equal. Because electric field lines point away from positive charges, the two field vectors at the midpoint point in directly opposite directions. Taking vector superposition gives a net electric field of zero.

Step-by-Step Solution

1
Determine the distance from each charge to the midpoint.
The midpoint distance r=0.2 m2=0.1 mr = \frac{0.2\text{ m}}{2} = 0.1\text{ m}.
Electric field calculation requires the distance from the point charge to the point of evaluation.
2
Calculate the magnitude of the electric field due to one charge.
E=kQr2=9.0×109×2.0×10−6(0.1)2=1.8×106 N C−1E = \frac{k Q}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.1)^2} = 1.8 \times 10^6\text{ N C}^{-1}.
Electric field intensity magnitude is given by Coulomb's field formula E=kQr2E = \frac{k Q}{r^2}.
3
Apply vector addition to find the net electric field at the midpoint.
Enet=E1−E2=1.8×106−1.8×106=0 N C−1E_{\text{net}} = E_1 - E_2 = 1.8 \times 10^6 - 1.8 \times 10^6 = 0\text{ N C}^{-1}.
Electric field is a vector quantity. Since both charges are positive, the field vectors point away from each charge and act in opposite directions at the midpoint.

Key Concept

Vector superposition of electric fields
Rate this question

Topics

Electrostatics and Electric Charges

More questions from this topic