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Two point charges qA = 3 µC and qB = −3 µC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?
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Two point charges qA = 3 µC and qB = −3 µC are located 20 cm apart in ...
Solution:
Given:
Charge qA = 3 µC
Charge qB = -3 µC
Distance between the charges, AB = 20 cm
(a) Electric field at the midpoint O of the line AB joining the two charges.
The electric field at point O can be found using the principle of superposition. The electric field due to qA and qB at point O is given by the vector sum of the electric fields due to each charge individually.
The electric field due to a point charge q at a distance r from the charge is given by:
E = kq/r^2
where k is the Coulomb's constant (9 × 10^9 Nm^2/C^2)
Applying the principle of superposition, the electric field at point O due to charges qA and qB is given by:
E = E1 + E2
where E1 is the electric field due to charge qA and E2 is the electric field due to charge qB.
The magnitude of the electric field due to each charge at point O is given by:
E1 = kqA/r1^2
E2 = kqB/r2^2
where r1 and r2 are the distances between the charges and point O respectively.
Using the distance formula, we can find r1 and r2 as:
r1 = r2 = AB/2 = 10 cm
Substituting the values, we get:
E1 = kqA/r1^2 = (9 × 10^9) × (3 × 10^-6)/(0.1)^2 = 2.7 × 10^5 N/C (directed towards qA)
E2 = kqB/r2^2 = (9 × 10^9) × (-3 × 10^-6)/(0.1)^2 = -2.7 × 10^5 N/C (directed towards qB)
The net electric field at point O is the vector sum of E1 and E2:
E = E1 + E2 = 0
Therefore, the electric field at point O is zero.
(b) Force experienced by the test charge of magnitude 1.5 × 10−9 C placed at point O.
The force experienced by a test charge q0 in an electric field E is given by:
F = q0E
Substituting the values, we get:
F = (1.5 × 10^-9) × 0 = 0 N
Explanation:
When a test charge is placed at point O, it experiences zero force because the electric field at point O is zero. This is because the electric fields due to qA and qB at point O are equal in magnitude and opposite in direction, resulting in a net electric field of zero. Therefore, the test charge experiences no force when placed at point O.
Given:
Charge qA = 3 µC
Charge qB = -3 µC
Distance between the charges, AB = 20 cm
(a) Electric field at the midpoint O of the line AB joining the two charges.
The electric field at point O can be found using the principle of superposition. The electric field due to qA and qB at point O is given by the vector sum of the electric fields due to each charge individually.
The electric field due to a point charge q at a distance r from the charge is given by:
E = kq/r^2
where k is the Coulomb's constant (9 × 10^9 Nm^2/C^2)
Applying the principle of superposition, the electric field at point O due to charges qA and qB is given by:
E = E1 + E2
where E1 is the electric field due to charge qA and E2 is the electric field due to charge qB.
The magnitude of the electric field due to each charge at point O is given by:
E1 = kqA/r1^2
E2 = kqB/r2^2
where r1 and r2 are the distances between the charges and point O respectively.
Using the distance formula, we can find r1 and r2 as:
r1 = r2 = AB/2 = 10 cm
Substituting the values, we get:
E1 = kqA/r1^2 = (9 × 10^9) × (3 × 10^-6)/(0.1)^2 = 2.7 × 10^5 N/C (directed towards qA)
E2 = kqB/r2^2 = (9 × 10^9) × (-3 × 10^-6)/(0.1)^2 = -2.7 × 10^5 N/C (directed towards qB)
The net electric field at point O is the vector sum of E1 and E2:
E = E1 + E2 = 0
Therefore, the electric field at point O is zero.
(b) Force experienced by the test charge of magnitude 1.5 × 10−9 C placed at point O.
The force experienced by a test charge q0 in an electric field E is given by:
F = q0E
Substituting the values, we get:
F = (1.5 × 10^-9) × 0 = 0 N
Explanation:
When a test charge is placed at point O, it experiences zero force because the electric field at point O is zero. This is because the electric fields due to qA and qB at point O are equal in magnitude and opposite in direction, resulting in a net electric field of zero. Therefore, the test charge experiences no force when placed at point O.
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Two point charges qA = 3 µC and qB = −3 µC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?
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Two point charges qA = 3 µC and qB = −3 µC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge? 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 qA = 3 µC and qB = −3 µC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two point charges qA = 3 µC and qB = −3 µC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?.
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