Question 91: Electrostats & Capacitance

Question

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

Accepted Answer

To determine the electric field from the given potential function, we need to use the relationship between electric field $$\mathbf{E}$$ and electric potential $$\mathrm{V}$$. The electric field is the negative gradient of the potential:
$$\mathbf{E} = -
abla \mathrm{V}$$
Given the potential function $$\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{xy}-\mathrm{y}+2 \mathrm{yz}$$, we need to compute the partial derivatives of $$\mathrm{V}$$ with respect to $$\mathrm{x}$$, $$\mathrm{y}$$, and $$\mathrm{z}$$.
First, calculate the partial derivative with respect to $$\mathrm{x}$$:
$$\frac{\partial \mathrm{V}}{\partial \mathrm{x}} = \frac{\partial}{\partial \mathrm{x}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 6 \mathrm{y}$$
Next, calculate the partial derivative with respect to $$\mathrm{y}$$:
$$\frac{\partial \mathrm{V}}{\partial \mathrm{y}} = \frac{\partial}{\partial \mathrm{y}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 6 \mathrm{x} - 1 + 2 \mathrm{z}$$
Finally, calculate the partial derivative with respect to $$\mathrm{z}$$:
$$\frac{\partial \mathrm{V}}{\partial \mathrm{z}} = \frac{\partial}{\partial \mathrm{z}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 2 \mathrm{y}$$
Now, the components of the electric field are:
$$ \mathbf{E}_x = -\frac{\partial \mathrm{V}}{\partial \mathrm{x}} = -6 \mathrm{y} $$
$$ \mathbf{E}_y = -\frac{\partial \mathrm{V}}{\partial \mathrm{y}} = -(6 \mathrm{x} - 1 + 2 \mathrm{z}) $$
$$ \mathbf{E}_z = -\frac{\partial \mathrm{V}}{\partial \mathrm{z}} = -2 \mathrm{y} $$
We are asked to find the electric field at the point $$(1,0,1)$$.
Substitute $$\mathrm{x} = 1$$, $$\mathrm{y} = 0$$, and $$\mathrm{z} = 1$$ into the components of the electric field:
$$ \mathbf{E}_x(1,0,1) = -6(0) = 0 $$
$$ \mathbf{E}_y(1,0,1) = -(6(1) - 1 + 2(1)) = -(6 - 1 + 2) = -(7) = -7 $$
$$ \mathbf{E}_z(1,0,1) = -2(0) = 0 $$
Therefore, the electric field at the point $$(1,0,1)$$ is:
$$\mathbf{E} = 0 \mathbf{i} - 7 \mathbf{j} + 0 \mathbf{k} = -7 \mathbf{j}$$
Hence, the correct option is:
Option A: $$\mathrm{-7 j}$$

Answer

$$\mathrm{-7 j}$$

Answer

$$+7 \mathrm{j}$$

Answer

$$\mathrm{-6 i+7 j}$$

Answer

$$\mathrm{6 i-7 j}$$

Electrostats & Capacitance

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

Choose the correct answer: