2 charges Q and –Q are kept at a distance 2a from each other. The electric field at theirmidpoint is E. At what distance from the midpoint on the perpendicular bisectorwould the electric field reduce to E/2?
Question
2 charges Q and –Q are kept at a distance 2a from each other. The electric field at theirmidpoint is E. At what distance from the midpoint on the perpendicular bisectorwould the electric field reduce to E/2?
Solution
The electric field at a point due to a charge Q is given by E = kQ/r², where r is the distance from the charge and k is Coulomb's constant.
At the midpoint between the two charges, the electric fields due to each charge are in opposite directions. Since the charges are equal in magnitude, the electric fields cancel each other out, so the net electric field E at the midpoint is zero.
We want to find the distance x from the midpoint on the perpendicular bisector where the electric field is E/2. At this point, the electric fields due to each charge do not cancel out completely.
The distance from each charge to the point is √(a² + x²) (by Pythagoras' theorem). The electric field at the point due to each charge is E' = kQ/√(a² + x²)².
The net electric field at the point is the vector sum of the electric fields due to each charge. Since the electric fields are at right angles to each other, we can use Pythagoras' theorem again to find the net electric field:
E = √(E'² + E'²) = √2E'
We want this to be E/2, so we set √2E' = E/2 and solve for x:
√2kQ/√(a² + x²)² = kQ/2a²
Squaring both sides and simplifying, we get
2k²Q²/(a² + x²) = k²Q²/4a²
Cross-multiplying and simplifying, we get
8a^4 = a² + x²
Subtracting a² from both sides, we get
7a² = x²
So x = √7a. This is the distance from the midpoint on the perpendicular bisector where the electric field is E/2.
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