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3 years ago

A uniform electric field is oriented in the −z direction. The magnitude of the electric field is 6500 N/C.

Physics

1 answer:

nignag [31]3 years ago

5 0

Answer:

In a plane parallel to the xy plane.

Explanation:

Direction of uniform electric field is oriented in the −z direction.

a) We know that the direction of equipotential surfaces are always normal to electric field direction. So, equipotential surface will be in a plane perpendicular to electric field that in a plane perpendicular to z-axis that is parallel to the xy plane.

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$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

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$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

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