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Electric potential at a point (x, y, z) in vacuum is V = 3x2 volt. Find the electric field intensity at the point (1m, 2m, 3m) .
- a)1 V/m
- b)3 V/m
- c)6 V/m
- d)2 V/m
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Electric potential at a point (x, y, z) in vacuum is V = 3x2 volt. Fin...
Answer:
To find the electric field intensity at a given point, you need to differentiate the electric potential with respect to each coordinate and then take the negative gradient of the potential.
Given that the electric potential at a point (x, y, z) is V = 3x^2 volts, we can find the electric field intensity at the point (1m, 2m, 3m) by taking the negative gradient of the potential.
Calculating the electric field intensity:
To find the electric field intensity, we need to calculate the partial derivatives of the potential with respect to each coordinate.
- Partial derivative with respect to x: ∂V/∂x = 6x
- Partial derivative with respect to y: ∂V/∂y = 0 (since V does not depend on y)
- Partial derivative with respect to z: ∂V/∂z = 0 (since V does not depend on z)
Now, we can calculate the electric field intensity at the given point (1m, 2m, 3m) by substituting the coordinates into the partial derivatives:
- Electric field intensity in the x-direction: Ex = -∂V/∂x = -6(1) = -6 V/m
- Electric field intensity in the y-direction: Ey = -∂V/∂y = 0
- Electric field intensity in the z-direction: Ez = -∂V/∂z = 0
The electric field intensity at the point (1m, 2m, 3m) is given by the vector sum of the components:
E = sqrt(Ex^2 + Ey^2 + Ez^2)
= sqrt((-6 V/m)^2 + (0)^2 + (0)^2)
= sqrt(36 V^2/m^2)
= 6 V/m
Therefore, the correct answer is option 'C' - 6 V/m.
To find the electric field intensity at a given point, you need to differentiate the electric potential with respect to each coordinate and then take the negative gradient of the potential.
Given that the electric potential at a point (x, y, z) is V = 3x^2 volts, we can find the electric field intensity at the point (1m, 2m, 3m) by taking the negative gradient of the potential.
Calculating the electric field intensity:
To find the electric field intensity, we need to calculate the partial derivatives of the potential with respect to each coordinate.
- Partial derivative with respect to x: ∂V/∂x = 6x
- Partial derivative with respect to y: ∂V/∂y = 0 (since V does not depend on y)
- Partial derivative with respect to z: ∂V/∂z = 0 (since V does not depend on z)
Now, we can calculate the electric field intensity at the given point (1m, 2m, 3m) by substituting the coordinates into the partial derivatives:
- Electric field intensity in the x-direction: Ex = -∂V/∂x = -6(1) = -6 V/m
- Electric field intensity in the y-direction: Ey = -∂V/∂y = 0
- Electric field intensity in the z-direction: Ez = -∂V/∂z = 0
The electric field intensity at the point (1m, 2m, 3m) is given by the vector sum of the components:
E = sqrt(Ex^2 + Ey^2 + Ez^2)
= sqrt((-6 V/m)^2 + (0)^2 + (0)^2)
= sqrt(36 V^2/m^2)
= 6 V/m
Therefore, the correct answer is option 'C' - 6 V/m.
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Community Answer
Electric potential at a point (x, y, z) in vacuum is V = 3x2 volt. Fin...
We know that
Here x = 1 m
Hence, E = 6 V/m.
Here x = 1 m
Hence, E = 6 V/m.
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Electric potential at a point (x, y, z) in vacuum is V = 3x2 volt. Find the electric field intensity at the point (1m, 2m, 3m) .a)1 V/mb)3 V/mc)6 V/md)2 V/mCorrect answer is option 'C'. Can you explain this answer?
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