Two point charges, [math: q1=+7μC] and [math: q2=−7μC] , are placed 0.21 m apart in a vacuum. What is the magnitude of the net electric field, at the midpoint between the two charges?*Note: [math: 1.23×104NC−1] is written as 1.23e4 N/C

Question

Two point charges, [math: q1=+7μC] and [math: q2=−7μC] , are placed 0.21 m apart in a vacuum. What is the magnitude of the net electric field, at the midpoint between the two charges?*Note: [math: 1.23×104NC−1] is written as 1.23e4 N/C

BlackTom AI · Accepted Answer

Electric field from a point charge is E = k|q|/r^2, so use Coulomb’s law and superposition. Here k = 8.988×10^9 N·m^2/C^2, q1 = +7.0×10^−6 C, q2 = −7.0×10^−6 C, separation d = 0.21 m. The midpoint is r = d/2 = 0.105 m from each charge.

Step-by-step calculation:
1. Distance squared: r^2 = (0.105 m)^2 = 0.011025 m^2.
2. Magnitude of the field from one charge:
   E_single = k|q|/r^2
            = (8.988×10^9 N·m^2/C^2)(7.0×10^−6 C) / 0.011025 m^2.
   Multiply numerator: 8.988×10^9 × 7.0×10^−6 = (8.988×7.0)×10^3 = 62.916×10^3 = 6.2916×10^4.
   Divide by r^2: E_single = 6.2916×10^4 / 0.011025 ≈ 5.7066×10^6 N/C.
3. Direction and superposition: The positive charge produces a field pointing away from it; the negative charge produces a field pointing toward it. At the midpoint the two fields point in the same direction (from + toward −), so magnitudes add.
4. Total field magnitude: E_net = 2 × E_single ≈ 2 × 5.7066×10^6 N/C ≈ 1.1413×10^7 N/C ≈ 1.14e7 N/C.

Why this option is correct:
- The calculation uses correct Coulomb constant, correct distances, and correctly adds the two equal-magnitude fields because their directions at the midpoint coincide. That yields 1.14×10^7 N/C, matching choice c.

Why other options are incorrect:
- (a) 2.85e6 N/C is about half of E_single (incorrect superposition or wrong r).
- (b) 2.85e12 N/C is many orders of magnitude too large (numerical error).
- (d) 5.71e6 N/C equals the field from one charge only, not the net from both.

Conclusion: The net electric field at the midpoint is 1.14e7 N/C, confirming answer c.

Two point charges, [math: q1=+7μC] and [math: q2=−7μC] , are placed 0.21 m apart in a vacuum. What is the magnitude of the net electric field, at the midpoint between the two charges?*Note: [math: 1.23×104NC−1] is written as 1.23e4 N/CSingle choice

Similar Questions

The potential difference between the plates of a parallel plate capacitor with the plate separation of 6 cm is 60 V. What is the electric field between the plates of this capacitor?

Calculate the electric field strength when a force of 4.0 × 10-9 𝑁 acts on a charge of 8.5 × 10-6𝐶.

Which, among the following is the correct expression for an electric field?

An electric field is:

The potential at x = 8 m is -2000 V and at x = 2 m is +400 V. What is the magnitude and direction of the electric field?

Two point charges are placed along a horizontal axis with the following values and positions: +3.0 µC at x = 0 cm and -6.0 µC at x = 20 cm. At what point along the x axis is the electric field zero?

The average distance of the electron from the proton in the hydrogen atom is 0.51 x 10-10 m. What is the electric field from the proton’s charge at the location of the electron? (ke = 8.99 x 109 N×m2/C2, elementary charge e = 1.6 x 10-19 C)

Consider the following diagram. Two parallel plates are separated in vacuum by 0.200 m. The top plate has a potential of 400 V and the bottom plate has a potential of −600 V. An alpha particle is released from the top plate.The acceleration of the alpha particle between the plates is

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler