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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

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

संबंधित प्रश्न

The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)


What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
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(i)
(ii)
(iii)
(iv)

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