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Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E, then the resultant electric field intensity at centre of square will be
- a)Zero
- b)4E
- c)E
- d)1/2E
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Four equal but like charge are placed at four corners of a square. The...
As the components of electric field intensity at diagonal are equal in magnitude and opposite in direction, thus the intensity of electric field will be zero.
Hence the correct answer would be option A.
Hence the correct answer would be option A.
Most Upvoted Answer
Four equal but like charge are placed at four corners of a square. The...
Given Information
Four equal charges, each having the magnitude q, are placed at the four corners of a square.
The distance between the charges and the center of the square is r.
The electric field intensity at the center of the square due to any one charge is E.
To find: The resultant electric field intensity at the center of the square.
Solution
The electric field intensity due to a point charge at a distance r from it is given by Coulomb's law as:
E = kq/r²
where k is the Coulomb's constant.
Electric Field Intensity at the Center of the Square due to Each Charge
Let's consider one of the charges, say Q1, and find the electric field intensity at the center of the square due to it.
As the distance between the charge and the center of the square is r, the electric field intensity at the center of the square due to Q1 is given by:
E1 = kQ1/r²
As all charges are equal and placed at the same distance from the center of the square, the electric field intensity due to each charge at the center of the square is E.
Therefore, the total electric field intensity at the center of the square due to all four charges is:
Etotal = E + E + E + E
= 4E
Resultant Electric Field Intensity at the Center of the Square
As all charges are of the same sign, the electric field vectors due to each charge add up vectorially at the center of the square.
Since the electric field intensity due to each charge is in the same direction, the resultant electric field intensity at the center of the square is given by the vector sum of the individual electric field intensities.
As all electric field vectors are equal in magnitude and direction, the resultant electric field intensity at the center of the square is zero.
Therefore, the correct option is (a) Zero.
Conclusion
The resultant electric field intensity at the center of the square due to four equal but like charges placed at its corners is zero.
Four equal charges, each having the magnitude q, are placed at the four corners of a square.
The distance between the charges and the center of the square is r.
The electric field intensity at the center of the square due to any one charge is E.
To find: The resultant electric field intensity at the center of the square.
Solution
The electric field intensity due to a point charge at a distance r from it is given by Coulomb's law as:
E = kq/r²
where k is the Coulomb's constant.
Electric Field Intensity at the Center of the Square due to Each Charge
Let's consider one of the charges, say Q1, and find the electric field intensity at the center of the square due to it.
As the distance between the charge and the center of the square is r, the electric field intensity at the center of the square due to Q1 is given by:
E1 = kQ1/r²
As all charges are equal and placed at the same distance from the center of the square, the electric field intensity due to each charge at the center of the square is E.
Therefore, the total electric field intensity at the center of the square due to all four charges is:
Etotal = E + E + E + E
= 4E
Resultant Electric Field Intensity at the Center of the Square
As all charges are of the same sign, the electric field vectors due to each charge add up vectorially at the center of the square.
Since the electric field intensity due to each charge is in the same direction, the resultant electric field intensity at the center of the square is given by the vector sum of the individual electric field intensities.
As all electric field vectors are equal in magnitude and direction, the resultant electric field intensity at the center of the square is zero.
Therefore, the correct option is (a) Zero.
Conclusion
The resultant electric field intensity at the center of the square due to four equal but like charges placed at its corners is zero.
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Community Answer
Four equal but like charge are placed at four corners of a square. The...
Answer is a) Zero because, the electric field is a vector quantity, so at the centre of square the electric field due to opposite charges (placed on the opposite corner of the square) cancel each other
and hence resultant electric field become zero.
and hence resultant electric field become zero.
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