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The electric potential varies in space according to the relation V = 3x + 4y. A particle of mass 10kg starts from rest from point (2,3.2)m under the influence of this field. Find the velocity of the particle in units of 10–3m/s, when it crosses the x-axis. The charge on the particle is 1μC. Assume V(x,y) are in SI units.
Correct answer is '2'. Can you explain this answer?
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The electric potential varies in space according to the relationV = 3x...
When particle crosses x-axis, y = 0
Initial y-coordinate was 3.2 m and 
At this instant x-coordinate will be,
The correct answer is: 2
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The electric potential varies in space according to the relationV = 3x...
To find the velocity of the particle, we need to find the electric field at point (2, 3.2) and then use it to calculate the particle's acceleration. Finally, we can use the acceleration to find the particle's velocity.
The electric field (E) is the negative gradient of the electric potential (V). Therefore, we need to find the partial derivatives of V with respect to x and y to find the components of the electric field.
∂V/∂x = 3
∂V/∂y = -4
So, the electric field at point (2, 3.2) is E = (3, -4).
The acceleration of the particle in the electric field is given by the equation a = qE/m, where q is the charge of the particle (not given) and m is its mass. Since the particle starts from rest, we can assume it has no charge (q = 0).
Therefore, the acceleration of the particle is a = 0.
Since the acceleration is zero, the velocity of the particle remains constant. But since the particle starts from rest, its velocity is also zero.
So, the velocity of the particle is 0 units of 10.
The electric field (E) is the negative gradient of the electric potential (V). Therefore, we need to find the partial derivatives of V with respect to x and y to find the components of the electric field.
∂V/∂x = 3
∂V/∂y = -4
So, the electric field at point (2, 3.2) is E = (3, -4).
The acceleration of the particle in the electric field is given by the equation a = qE/m, where q is the charge of the particle (not given) and m is its mass. Since the particle starts from rest, we can assume it has no charge (q = 0).
Therefore, the acceleration of the particle is a = 0.
Since the acceleration is zero, the velocity of the particle remains constant. But since the particle starts from rest, its velocity is also zero.
So, the velocity of the particle is 0 units of 10.
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The electric potential varies in space according to the relationV = 3x + 4y. A particle of mass10kgstarts from rest from point(2,3.2)munder the influence of this field. Find the velocity of the particle in units of10–3m/s, when it crosses thex-axis. The charge on the particle is1μC. AssumeV(x,y)are in SI units.Correct answer is '2'. Can you explain this answer?
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