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Chapter 11: Problem 1633

The electric potential \(\mathrm{V}\) at any point $\mathrm{x}, \mathrm{y}, \mathrm{z}\( (all in meter) in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}$ volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{Vm}^{-1}\) is \((\mathrm{A})+8 \mathrm{i} \wedge\) (B) \(-8 \mathrm{i} \wedge\) (C) \(-16 \mathrm{i}\) (D) \(+16 \mathrm{i}\)

Short Answer

Expert verified
The electric field at point \((1\ \mathrm{m}, 0, 2\ \mathrm{m})\) is \((B)\ -8\vec{i}\ \mathrm{Vm}^{-1}\).

Step by step solution

01

Find the derivative of the potential function with respect to x

Given \(V = 4x^2\) (y and z do not affect V). We will differentiate V with respect to x: \[ \frac{\partial V}{\partial x} = \frac{d(4x^2)}{dx} = 8x \]
02

Calculate the x-component of the electric field Ex using the formula E = -∇V

The x-component of the electric field Ex is given by: \[ Ex = -\frac{\partial V}{\partial x} = -8x \]
03

Evaluate Ex at the given point (1m, 0, 2m)

To find the electric field at the point (1m, 0, 2m), we plug in x=1 into Ex: \[ Ex = -8(1) = -8\ \mathrm{Vm}^{-1} \] So, the electric field at the given point is -8 Vm^{-1} in the x-direction. Checking the given options, we can conclude that the correct answer is: (B) \(-8\vec{i}\ \mathrm{Vm}^{-1}\)

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Most popular questions from this chapter

A parallel plate air capacitor has a capacitance \(18 \mu \mathrm{F}\). If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes \(72 \mu \mathrm{F}\). The dielectric constant of the medium is (A) 4 (B) 12 (C) 9 (D) 2

A simple pendulum consists of a small sphere of mass \(\mathrm{m}\) suspended by a thread of length \(\ell\). The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength \(\mathrm{E}\) directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is (A) $\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}-(\mathrm{q} \mathrm{E} / \mathrm{m})\\}]^{(1 / 2)}$ (B) \(\mathrm{T}=2 \pi(\ell / \mathrm{g})^{(1 / 2)}\) \(\left.\left.\left.\mathrm{m}_{\mathrm{}}\right\\}\right\\}\right]^{(1 / 2)}\) (D) \(\mathrm{T}=2 \pi[(\mathrm{m} \ell / \mathrm{qE})]^{(1 / 2)}\) (C) \(\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}+(\mathrm{qE} / \mathrm{t}\)

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)

A charge \(\mathrm{Q}\) is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are \(\ldots \ldots\) (A) \((Q / 3),(Q / 3)\) (B) \((\mathrm{Q} / 2),(\mathrm{Q} / 2)\) (C) \((Q / 4),(3 Q / 4)\) (D) \((Q / 5),(4 Q / 5)\)

An electric circuit requires a total capacitance of \(2 \mu \mathrm{F}\) across a potential of \(1000 \mathrm{~V}\). Large number of \(1 \mu \mathrm{F}\) capacitances are available each of which would breakdown if the potential is more then \(350 \mathrm{~V}\). How many capacitances are required to make the circuit? (A) 24 (B) 12 (C) 20 (D) 18
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