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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

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

A variable condenser is permanently connected to a \(100 \mathrm{~V}\) battery. If capacitor is changed from \(2 \mu \mathrm{F}\) to \(10 \mu \mathrm{F}\). then energy changes is equal to (A) \(2 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(6.5 \times 10^{-2} \mathrm{~J}\) (D) \(4 \times 10^{-2} \mathrm{~J}\)

An oil drop of 12 excess electrons is held stationary under a constant electric field of \(2.55 \times 10^{4} \mathrm{Vm}^{-1}\). If the density of the oil is \(1.26 \mathrm{gm} / \mathrm{cm}^{3}\) then the radius of the drop is \(\ldots \ldots \ldots \mathrm{m}\). (A) \(9.75 \times 10^{-7}\) (B) \(9.29 \times 10^{-7}\) (C) \(9.38 \times 10^{-8}\) (D) \(9.34 \times 10^{-8}\)

If 3 charges are placed at the vertices of equilateral triangle of charge ' \(q\) ' each. What is the net potential energy, if the side of equilateral triangle is \(\ell \mathrm{cm}\). (A) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(3 q^{2} / \ell\right)$ (B) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(2 q^{2} / \ell\right)$ (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(q^{2} / \ell\right)\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\left(4 q^{2} / \ell\right)$
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