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The electric potential at a point (x, y) in the xy-plane is given by: V = -kxy. The electric field intensity at a distance r from the origin varies as
- a)r
- b)2r
- c)r2
- d)2r2
Correct answer is option 'A'. Can you explain this answer?
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The electric potential at a point (x, y) in the xy-plane is given by: ...
Explanation:
Electric Potential (V) = -kxy
The electric field intensity (E) at a point is the negative gradient of the electric potential (V).
E = -∇V
∇ is the nabla operator which is defined as,
∇ = i(∂/∂x) + j(∂/∂y) + k(∂/∂z)
where i, j, and k are the unit vectors of x, y, and z directions.
So,
E = -∇V = -i(∂V/∂x) - j(∂V/∂y)
= -i(-ky) - j(-kx) [as V = -kxy]
= kxi + kyj
Magnitude of E, |E| = √(Ex^2 + Ey^2)
At a distance r = √(x^2 + y^2) from the origin,
|E| = √[(kx)^2 + (ky)^2] = k√(x^2 + y^2) = kr
Hence, the electric field intensity at a distance r from the origin varies as r.
Answer: (a) r
Electric Potential (V) = -kxy
The electric field intensity (E) at a point is the negative gradient of the electric potential (V).
E = -∇V
∇ is the nabla operator which is defined as,
∇ = i(∂/∂x) + j(∂/∂y) + k(∂/∂z)
where i, j, and k are the unit vectors of x, y, and z directions.
So,
E = -∇V = -i(∂V/∂x) - j(∂V/∂y)
= -i(-ky) - j(-kx) [as V = -kxy]
= kxi + kyj
Magnitude of E, |E| = √(Ex^2 + Ey^2)
At a distance r = √(x^2 + y^2) from the origin,
|E| = √[(kx)^2 + (ky)^2] = k√(x^2 + y^2) = kr
Hence, the electric field intensity at a distance r from the origin varies as r.
Answer: (a) r
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