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Chapter 11: Problem 1571

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

Expert verified
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

01

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: $$ \nabla V = (8x, 0, 0) $$
02

Evaluate the gradient at the given point

Now, we will evaluate the gradient of the electric potential at the given point (1m, 0.2m): $$ \nabla V(1, 0.2) = (8(1), 0, 0) = (8, 0, 0) $$
03

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: $$ \textbf{E}(1, 0.2) = -(8, 0, 0) = (-8, 0, 0) \; \text{V/m} $$
04

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

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Most popular questions from this chapter

Electric charges of \(+10 \mu \mathrm{c}, 5 \mu \mathrm{c},-3 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed at the corners of a square of side $\sqrt{2 m}\( the potential at the centre of the square is \)\ldots \ldots$ (A) \(1.8 \mathrm{~V}\) (B) \(1.8 \times 10^{5} \mathrm{~V}\) (C) \(1.8 \times 10^{6} \mathrm{~V}\) (D) \(1.8 \times 10^{4} \mathrm{~V}\)

A parallel plate condenser with dielectric of constant \(\mathrm{K}\) between the plates has a capacity \(\mathrm{C}\) and is charged to potential \(\mathrm{v}\) volt. The dielectric slab is slowly removed from between the plates and reinserted. The net work done by the system in this process is (A) Zero (B) \((1 / 2)(\mathrm{K}-1) \mathrm{cv}^{2}\) (C) \((\mathrm{K}-1) \mathrm{cv}^{2}\) (D) \(\mathrm{cv}^{2}[(\mathrm{~K}-1) / \mathrm{k}]\)

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)

The electric potential \(\mathrm{V}\) is given as a function of distance \(\mathrm{x}\) (meter) by $\mathrm{V}=\left(5 \mathrm{x}^{2}+10 \mathrm{x}-9\right)\( volt. Value of electric field at \)\mathrm{x}=1$ is \(\ldots \ldots\) \((\mathrm{A})-20(\mathrm{v} / \mathrm{m})\) (B) \(6(\mathrm{v} / \mathrm{m})\) (C) \(11(\mathrm{v} / \mathrm{m})\) (D) \(-23(\mathrm{v} / \mathrm{m})\)

A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}\(. The separation between its plates is \)\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectric constant \(\mathrm{K}_{1}=3\) and thickness \(\mathrm{d} / 3\) while the other one has dielectric constant \(\mathrm{K}_{2}=6\) and thickness \(2 \mathrm{~d} / 3\). Capacitance of the capacitor is now (A) \(1.8 \mathrm{pF}\) (B) \(20.25 \mathrm{pF}\) (C) \(40.5 \mathrm{pF}\) (D) \(45 \mathrm{pF}\)
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