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An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ is
- a)π/3
- b)π/3 + tan-1√3/2
- c)3π/2
- d)tan-1√3/2
Correct answer is option 'B'. Can you explain this answer?
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An electric dipole is placed along the X-axis at the origin O. A point...
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An electric dipole is placed along the X-axis at the origin O. A point...
Let's assume that the positive direction along the X-axis is to the right, and the negative direction along the X-axis is to the left.
The electric dipole consists of two equal and opposite charges, one positive (+q) and one negative (-q), separated by a distance d. The dipole moment (p) is defined as the product of the charge magnitude and the separation distance:
p = q * d
In this case, we can assume that the positive charge (+q) is located at position (d/2, 0, 0), and the negative charge (-q) is located at position (-d/2, 0, 0).
The point P is located at a distance of 20 cm from the origin O, which is at (0, 0, 0).
Let's assume that the line connecting O and P makes an angle θ with the positive X-axis. The coordinates of point P can be written as:
P = (20 * cos(θ), 20 * sin(θ), 0)
To calculate the electric field at point P due to the electric dipole, we can use the formula for the electric field due to a point charge:
E = (k * q) / r^2
Where E is the electric field, k is the electrostatic constant (k = 9 * 10^9 Nm^2/C^2), q is the charge magnitude, and r is the distance between the charge and the point where the electric field is being calculated.
The electric field due to the positive charge (+q) at point P can be calculated as:
E_positive = (k * q) / r_positive^2
Where r_positive is the distance between the positive charge and point P:
r_positive = sqrt((20 * cos(θ) - d/2)^2 + (20 * sin(θ))^2)
Similarly, the electric field due to the negative charge (-q) at point P can be calculated as:
E_negative = (k * q) / r_negative^2
Where r_negative is the distance between the negative charge and point P:
r_negative = sqrt((20 * cos(θ) + d/2)^2 + (20 * sin(θ))^2)
The total electric field at point P due to the electric dipole is the vector sum of the electric fields due to the positive and negative charges:
E_total = E_positive - E_negative
Now, you can substitute the values of q, d, θ, and the distance 20 cm into these equations to calculate the electric field at point P.
The electric dipole consists of two equal and opposite charges, one positive (+q) and one negative (-q), separated by a distance d. The dipole moment (p) is defined as the product of the charge magnitude and the separation distance:
p = q * d
In this case, we can assume that the positive charge (+q) is located at position (d/2, 0, 0), and the negative charge (-q) is located at position (-d/2, 0, 0).
The point P is located at a distance of 20 cm from the origin O, which is at (0, 0, 0).
Let's assume that the line connecting O and P makes an angle θ with the positive X-axis. The coordinates of point P can be written as:
P = (20 * cos(θ), 20 * sin(θ), 0)
To calculate the electric field at point P due to the electric dipole, we can use the formula for the electric field due to a point charge:
E = (k * q) / r^2
Where E is the electric field, k is the electrostatic constant (k = 9 * 10^9 Nm^2/C^2), q is the charge magnitude, and r is the distance between the charge and the point where the electric field is being calculated.
The electric field due to the positive charge (+q) at point P can be calculated as:
E_positive = (k * q) / r_positive^2
Where r_positive is the distance between the positive charge and point P:
r_positive = sqrt((20 * cos(θ) - d/2)^2 + (20 * sin(θ))^2)
Similarly, the electric field due to the negative charge (-q) at point P can be calculated as:
E_negative = (k * q) / r_negative^2
Where r_negative is the distance between the negative charge and point P:
r_negative = sqrt((20 * cos(θ) + d/2)^2 + (20 * sin(θ))^2)
The total electric field at point P due to the electric dipole is the vector sum of the electric fields due to the positive and negative charges:
E_total = E_positive - E_negative
Now, you can substitute the values of q, d, θ, and the distance 20 cm into these equations to calculate the electric field at point P.
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An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer?
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An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer?.
An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric dipole is placed along the X-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the X axis. If electric field at P makes an angle θ with X-axis, the value of θ isa)π/3b)π/3 + tan-1√3/2c)3π/2d)tan-1√3/2Correct answer is option 'B'. Can you explain this answer?.
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