Question

If potential (in volts) in a region is expressed as $V(x, y , z)$ = $6xy - y + 2yz$, the electric field $(in\, N/C)$ at point $(1, 1, 0)$ is

• A. $ -(2\widehat {i} + 3\widehat {j} + \widehat {k} )$
• B. $ -(6\widehat {i} + 9\widehat {j} + \widehat {k} )$
• C. $ -(3\widehat {i} + 5\widehat {j} + 3\widehat {k} )$
• D. $ -(6\widehat {i} + 5\widehat {j} + 2\widehat {k} )$

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the electric field at a given point based on the potential function V(x, y, z) = 6xy - y + 2yz. The electric field \mathbf{E} is related to the electric potential V by the negative gradient:

• \mathbf{E} = -\nabla V\

The gradient \nabla V is given by the partial derivatives:

• \nabla V = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)

Calculating each partial derivative:

• \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y.
• \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z.
• \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y.

Substitute the point (1, 1, 0) into the expressions for the partial derivatives:

• \frac{\partial V}{\partial x}\Big|_{(1, 1, 0)} = 6(1) = 6.
• \frac{\partial V}{\partial y}\Big|_{(1, 1, 0)} = 6(1) - 1 + 2(0) = 5.
• \frac{\partial V}{\partial z}\Big|_{(1, 1, 0)} = 2(1) = 2.

Thus, the gradient \nabla V at the point (1, 1, 0) is (6, 5, 2). Applying the negative, the electric field is:

• \mathbf{E} = - (6\widehat{i} + 5\widehat{j} + 2\widehat{k}).

Therefore, the electric field at the point (1, 1, 0) is -(6\widehat{i} + 5\widehat{j} + 2\widehat{k}), which matches the correct answer.