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NEET PHYSICSELECTRIC CHARGES AND FIELDSEasy

Question

Two point charges qA=3μCq_A = 3 \, \mu\text{C} and qB=3μCq_B = -3 \, \mu\text{C} are located 20cm20 \, \text{cm} apart in a vacuum. The electric field at the midpoint OO of the line ABAB joining the two charges is:

A

4.5×106N/C4.5 \times 10^6 \, \text{N/C} along OAOA

B

5.4×106N/C5.4 \times 10^6 \, \text{N/C} along OAOA

C

4.5×106N/C4.5 \times 10^6 \, \text{N/C} along OBOB

D

5.4×106N/C5.4 \times 10^6 \, \text{N/C} along OBOB

Step-by-Step Solution

The electric field EE due to a point charge is given by E=14πε0qr2E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}. Given: qA=3×106Cq_A = 3 \times 10^{-6} \, \text{C}, qB=3×106Cq_B = -3 \times 10^{-6} \, \text{C}. Distance AB=20cm=0.2mAB = 20 \, \text{cm} = 0.2 \, \text{m}. The midpoint OO is at a distance r=10cm=0.1mr = 10 \, \text{cm} = 0.1 \, \text{m} from both charges.

  1. Field due to positive charge qAq_A at OO points away from AA (towards BB). Magnitude EA=(9×109)(3×106)(0.1)2=2.7×106N/CE_A = \frac{(9 \times 10^9)(3 \times 10^{-6})}{(0.1)^2} = 2.7 \times 10^6 \, \text{N/C}.
  2. Field due to negative charge qBq_B at OO points towards BB. Magnitude EB=(9×109)(3×106)(0.1)2=2.7×106N/CE_B = \frac{(9 \times 10^9)(3 \times 10^{-6})}{(0.1)^2} = 2.7 \times 10^6 \, \text{N/C}. Since both fields point in the same direction (along OBOB), the net electric field is the sum of their magnitudes: Enet=EA+EB=(2.7+2.7)×106=5.4×106N/CE_{net} = E_A + E_B = (2.7 + 2.7) \times 10^6 = 5.4 \times 10^6 \, \text{N/C}, directed along OBOB.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTRIC CHARGES AND FIELDS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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