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Chapter 11: Problem 1576

The electric Potential at a point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( is given by \)\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{x} \mathrm{z}^{3}+4\(. The electric field \)\mathrm{E}^{\boldsymbol{T}}$ at that point is \(\ldots \ldots\) (A) $i \wedge\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\mathrm{j} \wedge \mathrm{x}^{2}+\mathrm{k} \wedge 3 \mathrm{xz}^{2}$ (B) $i \wedge 2 \mathrm{xy}+\mathrm{j} \wedge\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\mathrm{k} \wedge\left(3 \mathrm{xy}-\mathrm{y}^{2}\right)$ (C) \(i \wedge z^{3}+j \wedge x y z+k \wedge z^{2}\) (D) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)

Short Answer

Expert verified
The electric field at point P, \(\mathbf{E}(x, y, z) = \left(2xy + z^3, x^2, 3xz^2\right)\), corresponds to the answer (A) and can be represented as: \[\mathbf{E} = \bold{i}(2xy + z^3) + \bold{j}(x^2) + \bold{k}(3xz^2)\]

Step by step solution

01

Identify the electric potential function

Here, the electric potential function \(V(x, y, z)\) is given by: \[V(x, y, z) = -x^2y - xz^3 + 4\]
02

Calculate the gradient of V

To find the electric field, we need to compute the gradient of V, i.e., its derivative along \(x\), \(y\), and \(z\) directions. The gradient of V is given as: \(\nabla V =\left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)\). Now, let's compute each partial derivative: \(\frac{\partial V}{\partial x} = \frac{\partial(-x^2y - xz^3 + 4)}{\partial x} = -2xy - z^3\) \(\frac{\partial V}{\partial y} = \frac{\partial(-x^2y - xz^3 + 4)}{\partial y} = -x^2\) \(\frac{\partial V}{\partial z} = \frac{\partial(-x^2y - xz^3 + 4)}{\partial z} = -3xz^2\) So, the gradient of V is: \(\nabla V =\left(-2xy - z^3, -x^2, -3xz^2\right)\)
03

Determine the electric field using the gradient of V

We know that the electric field \(\mathbf{E}\) is given by: \[\mathbf{E} = -\nabla V\] Substituting the values of the gradient from step 2, we get: \[\mathbf{E} = -(-2xy - z^3, -x^2, -3xz^2) = \left(2xy + z^3, x^2, 3xz^2\right)\] So, the electric field at point P, \(\mathbf{E}(x, y, z) = \left(2xy + z^3, x^2, 3xz^2\right)\). This corresponds to the answer (A), which is: \[\mathbf{E} = \bold{i}(2xy + z^3) + \bold{j}(x^2) + \bold{k}(3xz^2)\]

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Most popular questions from this chapter

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 thin spherical conducting shell of radius \(\mathrm{R}\) has a charge q. Another charge \(Q\) is placed at the centre of the shell. The electrostatic potential at a point p a distance \((\mathrm{R} / 2)\) from the centre of the shell is ..... (A) \(\left[(q+Q) /\left(4 \pi \epsilon_{0}\right)\right](2 / R)\) (B) $\left[\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right\\}-\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right]\right.$ (C) $\left[\left\\{(2 Q) /\left(4 \pi \in_{0} R\right)\right\\}+\left\\{q /\left(4 \pi \epsilon_{0} R\right)\right]\right.$ (D) \(\left[(2 \mathrm{Q}) /\left(4 \pi \epsilon_{0} \mathrm{R}\right)\right]\)

N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of \(\mathrm{N}\) is \(\ldots \ldots\) (A) 4 (B) 6 (C) 8 (D) 10

Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)
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