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1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q will
- a)Execute simple harmonic motion about the origin
- b)Move to the origin and remain at rest
- c)Move to infinity
- d)Execute oscillatory but not simple harmonic motion
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
1) Two equal negative charge ? q are fixed at the fixed points (0,a) a...
By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O. Net force on charge Q 
\[{{F}_{net}}\Rightarrow 2F\cos \theta =2\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-qQ}{({{a}^{2}}+{{x}^{2}})}\times \frac{x}{{{({{a}^{2}}+{{x}^{2}})}^{{1}/{{}}\;2}}}\] i.e., \[{{F}_{net}}=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2qQx}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{{3}/{{}}\;2}}}\] As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmoni
\[{{F}_{net}}\Rightarrow 2F\cos \theta =2\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-qQ}{({{a}^{2}}+{{x}^{2}})}\times \frac{x}{{{({{a}^{2}}+{{x}^{2}})}^{{1}/{{}}\;2}}}\] i.e., \[{{F}_{net}}=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2qQx}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{{3}/{{}}\;2}}}\] As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmoniMost Upvoted Answer
1) Two equal negative charge ? q are fixed at the fixed points (0,a) a...
-a) on the y-axis. A positive charge ? Q is placed at the origin (0,0). Find the net electrostatic force on the positive charge.
To find the net electrostatic force on the positive charge, we need to calculate the force due to each of the negative charges and then add them up.
Let's first calculate the force due to the negative charge at (0,a). The distance between the charges is d = a, and the magnitude of the force is given by Coulomb's law:
F1 = k * Q * q / d^2
where k is the Coulomb constant, Q is the positive charge, q is the negative charge, and d is the distance between the charges.
Since the negative charge is on the y-axis, the force it exerts on the positive charge will also be along the y-axis. Therefore, we can write:
F1y = F1 * sin(theta)
where theta is the angle between the force and the y-axis. In this case, theta is 90 degrees, so sin(theta) = 1.
Putting it all together, we get:
F1y = k * Q * q / a^2
Now let's calculate the force due to the negative charge at (0,-a). The distance between the charges is also d = a, so the magnitude of the force is the same as before:
F2 = k * Q * q / a^2
However, this force is in the opposite direction to the force due to the first negative charge, so its y-component is:
F2y = -F2 * sin(theta) = -k * Q * q / a^2
Adding up the two forces, we get:
Fnet = F1y + F2y = k * Q * q / a^2 - k * Q * q / a^2 = 0
Therefore, the net electrostatic force on the positive charge is zero. This makes sense, because the two negative charges are symmetrically placed on either side of the positive charge, so their forces cancel out.
To find the net electrostatic force on the positive charge, we need to calculate the force due to each of the negative charges and then add them up.
Let's first calculate the force due to the negative charge at (0,a). The distance between the charges is d = a, and the magnitude of the force is given by Coulomb's law:
F1 = k * Q * q / d^2
where k is the Coulomb constant, Q is the positive charge, q is the negative charge, and d is the distance between the charges.
Since the negative charge is on the y-axis, the force it exerts on the positive charge will also be along the y-axis. Therefore, we can write:
F1y = F1 * sin(theta)
where theta is the angle between the force and the y-axis. In this case, theta is 90 degrees, so sin(theta) = 1.
Putting it all together, we get:
F1y = k * Q * q / a^2
Now let's calculate the force due to the negative charge at (0,-a). The distance between the charges is also d = a, so the magnitude of the force is the same as before:
F2 = k * Q * q / a^2
However, this force is in the opposite direction to the force due to the first negative charge, so its y-component is:
F2y = -F2 * sin(theta) = -k * Q * q / a^2
Adding up the two forces, we get:
Fnet = F1y + F2y = k * Q * q / a^2 - k * Q * q / a^2 = 0
Therefore, the net electrostatic force on the positive charge is zero. This makes sense, because the two negative charges are symmetrically placed on either side of the positive charge, so their forces cancel out.
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1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer?
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1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer?.
1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice 1) Two equal negative charge ? q are fixed at the fixed points (0,a) and (0,−a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. The charge Q willa)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice NEET tests.
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