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81
The capacitance between two plates increases with
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Answer: Option D
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82
The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be
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Answer: Option C
Solution:
Here's an explanation of the problem and solution:

The question deals with the electric field intensity due to a point charge.

The electric field intensity (E) due to a point charge (Q) at a distance (r) is given by the formula: E = kQ/r2, where k is a constant.

This means the electric field intensity is inversely proportional to the square of the distance.

So, if the distance is halved, the electric field intensity will be quadrupled (multiplied by 4).

Initially, at 4 meters, the electric field intensity is 200 N/C.

When the distance is reduced to 2 meters (which is half of 4 meters), the new electric field intensity will be 200 N/C * 4 = 800 N/C.

Therefore, the correct answer is Option C: 800 N/C
83
A mica capacitor and a ceramic capacitor both have the same physical dimensions. Which will have more value of capacitance ?
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Answer: Option A
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84
A unit tube of flux is known as tube
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Answer: Option B
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85
Out of the following capacitors of identical rating which one will have the smallest dimensions ?
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Answer: Option C
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86
Which of the following statements is incorrect ?
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Answer: Option B
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87
For a good 0.05 uF capacitor ohmmeter reading should
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Answer: Option A
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88
The total deficiency or excess of electrons in a body is known as
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Answer: Option D
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89
The electrostatic force between two charges of one coulomb each and placed at a distance of 0.5 m will be
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Answer: Option D
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90
Which of the following is not a vector ?
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Answer: Option D
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