Question

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

• A. \( -7j \)
• B. \( +7j \)
• C. \( -6i + 7j \)
• D. \( 6i - 7j \)

Hint:

To find the electric field from the potential, remember that the electric field is the negative gradient of the potential. This means you take the partial derivatives with respect to \(x\), \(y\), and \(z\), and multiply them by the unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) respectively.

Answer 1

The Correct Option is A

The electric field is the negative gradient of the potential. The gradient of the potential function in three dimensions is: \[ \vec{E} = - \nabla V = - \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \] We are given the potential: \[ V(x, y, z) = 6xy - y + 2yz \] Now, calculate the partial derivatives: 
1. \( \frac{\partial V}{\partial x} = 6y \) 2. \( \frac{\partial V}{\partial y} = 6x - 1 + 2z \) 3. \( \frac{\partial V}{\partial z} = 2y \) 
At the point \( (1, 0, 1) \): 1. \( \frac{\partial V}{\partial x} = 6(0) = 0 \) 
2. \( \frac{\partial V}{\partial y} = 6(1) - 1 + 2(1) = 6 - 1 + 2 = 7 \) 
3. \( \frac{\partial V}{\partial z} = 2(0) = 0 \) 
Thus, the electric field is: \[ \vec{E} = - \left( 0 \hat{i} + 7 \hat{j} + 0 \hat{k} \right) = -7 \hat{j} \] 
Therefore, the electric field at point \( (1, 0, 1) \) is \( -7j \).