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The electric field intensity at a point A due to a point charge is observed to be four times greater than that observed at another point B. If the charge remains unchanged, what must be the possible reason for the increase in field intensity at A?
1. The distance from the point charge to A is half of the distance between charge and point B
2. The distance from the point charge to A is 4 times more than that of B 
3. The charge got reduced to its one-fourth
4. none of the above

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Correct Answer - Option 1 : The distance from the point charge to A is half of the distance between charge and point B

The correct answer is option 1) i.e. The distance from the point charge to A is half of the distance between charge and point B

CONCEPT:

  • Electric field strength is a measure of the electric field intensity at a point due to a charge. It is the amount of electric force produced per unit charge.

Electric field strength (E) at a point due to a point charge q is given by the formula:

\(E =\frac{q}{4\pi \epsilon_0 r^2}\)
q4πϵ0r2

Where r is the distance from the point charge to the point at which E is determined.

EXPLANATION:

If the charge remains constant, 

\(E \propto \frac{1}{r^2}\)

So, if the electric field is four times greater at A,

\(E \propto4 \times \frac{1}{r^2} \Rightarrow E \propto \frac{1}{(r/2)^2}\)

  • Therefore, the distance from the point charge to A is half the distance between the charge and point B.

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