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
A. 400 N/C
B. 600 N/C
C. 800 N/C
D. 1200 N/C
Answer: Option C
Solution (By Examveda Team)
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
This Question Belongs to Electrical Engineering >> Electrostatics
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The electric field intensity due to a point charge is inversely proportional to the square of the distance. (E= rac{kQ}{r^{2}}), where (E) is the electric field intensity, (k) is Coulomb's constant, (Q) is the charge, and (r) is the distance.