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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

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प्रश्न

Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.

  1. Identify an equipotential surface of the system.
  2. What is the direction of the electric field at every point on this surface?
संख्यात्मक
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उत्तर

Given: q1 = 2 μC = 2 × 10−6 C

q2 = −2 µC = 2 × 10−6 C 

r = 6 cm = 0.06 m

(a) Potential will be zero due to both charges at the equipotential surface.

`1/(4piε_0)[q_1/x + q_2/((0.06 - x))] = 0`

`q_1/x = -q_2/((0.06 - x))`

`(2 xx 10^-6)/x = -((-2 xx 10^-6))/([(0.06) - x])`

x = 0.06 − x 

x + x = 0.06

2x = 0.06

`x = 0.06/2`

x = 0.03 m

i.e., the plane normal to AB and passing through its mid-point has zero potential everywhere. 

(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.

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अध्याय 2: Electrostatic Potential and Capacitance - EXERCISES [पृष्ठ ७९]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 2 Electrostatic Potential and Capacitance
EXERCISES | Q 2.3 | पृष्ठ ७९
एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 2 Electrostatic Potential and Capacitance
Exercise | Q 2.3 | पृष्ठ ८६

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