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Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ?
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Calculation of Electric Field Intensity
The electric field intensity at the middle point of the line joining two point charges can be calculated using the formula:
E = k*q/r^2
where k is the Coulomb's constant, q is the magnitude of the charge, and r is the distance between the charges.
Substitution of Values
In this case, the charges are of magnitude 0.01 microcoulomb and -0.01 microcoulomb, and they are placed 10 cm apart in vacuum. Therefore,
q1 = 0.01 microcoulomb
q2 = -0.01 microcoulomb
r = 10 cm = 0.1 m
k = 9 x 10^9 N m^2/C^2
Substituting these values in the formula, we get:
E = (9 x 10^9 N m^2/C^2) * (0.01 x 10^-6 C) / (0.05 m)^2
E = 3.6 x 10^4 N/C
Direction of Electric Field Intensity
The direction of the electric field intensity at the midpoint of the line joining the two charges can be determined using the principle of superposition. Since the charges have opposite signs, the electric field at the midpoint will be directed towards the negative charge and away from the positive charge. Therefore, the direction of the electric field intensity in this case is from the negative charge to the positive charge.
Conclusion
The magnitude of the electric field intensity at the midpoint of the line joining two point charges of magnitude 0.01 microcoulomb and -0.01 microcoulomb, placed 10 cm apart in vacuum, is 3.6 x 10^4 N/C, directed from the negative charge to the positive charge.
The electric field intensity at the middle point of the line joining two point charges can be calculated using the formula:
E = k*q/r^2
where k is the Coulomb's constant, q is the magnitude of the charge, and r is the distance between the charges.
Substitution of Values
In this case, the charges are of magnitude 0.01 microcoulomb and -0.01 microcoulomb, and they are placed 10 cm apart in vacuum. Therefore,
q1 = 0.01 microcoulomb
q2 = -0.01 microcoulomb
r = 10 cm = 0.1 m
k = 9 x 10^9 N m^2/C^2
Substituting these values in the formula, we get:
E = (9 x 10^9 N m^2/C^2) * (0.01 x 10^-6 C) / (0.05 m)^2
E = 3.6 x 10^4 N/C
Direction of Electric Field Intensity
The direction of the electric field intensity at the midpoint of the line joining the two charges can be determined using the principle of superposition. Since the charges have opposite signs, the electric field at the midpoint will be directed towards the negative charge and away from the positive charge. Therefore, the direction of the electric field intensity in this case is from the negative charge to the positive charge.
Conclusion
The magnitude of the electric field intensity at the midpoint of the line joining two point charges of magnitude 0.01 microcoulomb and -0.01 microcoulomb, placed 10 cm apart in vacuum, is 3.6 x 10^4 N/C, directed from the negative charge to the positive charge.
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Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ?
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Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ?.
Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two point charges 0.01micro column and -0.01microcoulomb are placed 10 cm apart in vacuum. calculated the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ?.
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