Question

The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

Short answer

The electric field at the point (1m, 0.2m) is \(-8 \mathrm{V/m}\), along the negative x-axis. The correct answer is (A).

Step-by-step solution

Find the gradient of the electric potential V

First, we will find the gradient of the given electric potential V(x, y, z) = 4x². The gradient is a vector that contains the partial derivatives of the function with respect to each coordinate: $$ \nabla V = \left(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\right) $$ Compute the partial derivatives: $$ \frac{\partial V}{\partial x} = 8x, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0 $$ So, the gradient of the electric potential is: $$ \nabla V = (8x, 0, 0) $$

Evaluate the gradient at the given point

Now, we will evaluate the gradient of the electric potential at the given point (1m, 0.2m): $$ \nabla V(1, 0.2) = (8(1), 0, 0) = (8, 0, 0) $$

Find the electric field

The electric field E is the negative gradient of the electric potential: $$ \textbf{E} = -\nabla V $$ So, at the point (1m, 0.2m), the electric field is: $$ \textbf{E}(1, 0.2) = -(8, 0, 0) = (-8, 0, 0) \; \text{V/m} $$

Match the answer with the given options

The electric field at the point (1m, 0.2m) is (-8, 0, 0) V/m, which means it has a magnitude of 8 V/m and points along the negative x-axis. This matches option (A) in the exercise. Therefore, the correct answer is: (A) 8, along negative x-axis