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Chapter 16: Problem 30

The electric potential \(V\) at any point \((x, y, z)\) (all in metres) in space is given by \(V=4 x^{2}\) volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in volt/metre is (a) 8 along negative \(x\) -axis (b) 8 along positive \(x\) -axis (c) 16 along negative \(x\) -axis (d) 16 along positive \(x\) -axis

Short Answer

Expert verified
(a) 8 along negative \(x\)-axis.

Step by step solution

01

Understand the Electric Field and Potential Relationship

The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the gradient of the potential. The relationship is given by \( \mathbf{E} = - abla V \), where \( abla V \) is the spatial gradient of the potential function \( V \).
02

Compute the Gradient of V

Find the partial derivatives of \( V \) with respect to \( x \), \( y \), and \( z \). For \( V = 4x^2 \), compute:1. \( \frac{\partial V}{\partial x} = 8x \)2. \( \frac{\partial V}{\partial y} = 0 \)3. \( \frac{\partial V}{\partial z} = 0 \)
03

Calculate the Electric Field Vector

Using the gradient found in Step 2, substitute into \( \mathbf{E} = -abla V \):\[ \mathbf{E} = - (8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} \]
04

Evaluate the Electric Field at the Given Point

Substitute the coordinates \((x, y, z) = (1, 0, 2)\) into \( -8x \hat{i} \):\[ \mathbf{E} = -8(1) \hat{i} = -8 \hat{i} \]
05

Determine the Direction of the Electric Field

The electric field \( \mathbf{E} = -8 \hat{i} \) means the electric field has a magnitude of 8 and points along the negative \( x \)-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
Electric potential, often denoted as \( V \), is a measure of the potential energy per unit charge at a specific location in an electric field. It's essentially a way to quantify how much work it would take to move a charge to that point from infinity. For this exercise, we're given an electric potential that depends only on the \( x \)-coordinate, \( V = 4x^2 \). This tells us how the potential changes as we move along the \( x \)-axis.
  • Units: The electric potential is measured in volts.
  • Link to energy: It is essentially the potential energy per unit charge.
  • Scalar Quantity: Unlike vectors, it does not have direction.
Understanding electric potential is key to moving forward with calculating the electric field, as changes in potential directly influence the field's properties.
Gradient
The concept of a gradient offers a bridge between scalar and vector fields in mathematics and physics. The gradient of a scalar function gives you a vector that points in the direction of the greatest increase of that function, with a magnitude that represents the rate of increase. In this exercise, the gradient of the potential \( V = 4x^2 \) focuses only on derivatives with respect to spatial variables.For the given potential, the gradient \( abla V \) is:
  1. \( \frac{\partial V}{\partial x} = 8x \): Indicates how \( V \) changes along the \( x \)-axis.
  2. \( \frac{\partial V}{\partial y} = 0 \)
  3. \( \frac{\partial V}{\partial z} = 0 \)
This shows that the potential only changes in the \( x \)-direction for this problem.
Partial Derivatives
Partial derivatives are utilized to understand how a function changes as just one variable is varied, keeping the others constant. When dealing with multivariable functions, partial derivatives help us compute the effect of changing each variable individually. In this case, we find the partial derivatives of our potential \( V = 4x^2 \):
  • \( \frac{\partial V}{\partial x} = 8x \): This derivative tells us how \( V \) changes with \( x \) and is crucial for finding \( abla V \).
  • \( \frac{\partial V}{\partial y} = 0 \) and \( \frac{\partial V}{\partial z} = 0 \): Since our potential does not change with \( y \) or \( z \), these derivatives are zero.
Computing these derivatives is an important step before forming the electric field vector.
Electric Field Vector
An electric field vector \( \mathbf{E} \) is a fundamental concept that encapsulates the influence a charged object exerts on other charges in its vicinity. It's a vector because it has both a magnitude and direction. The electric field \( \mathbf{E} \) is derived via the negative gradient of the electric potential: \( \mathbf{E} = - abla V \).After computing the gradient as \( 8x \hat{i} \), the electric field vector becomes:\[ \mathbf{E} = - (8x \hat{i}) = -8x \hat{i} \]
  • Magnitude: In this scenario, when evaluated at \( (1, 0, 2) \), \( \mathbf{E} = -8 \hat{i} \).
  • Direction: Points along the negative \( x \)-axis because of the negative sign.
This negatively directed vector indicates the direction that a positive test charge would naturally move if placed in the field.

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Most popular questions from this chapter

In Millikan's oil drop experiment a mist of oil is produced from an atomiser and tiny drops of oil are observed. If the oil drop is allowed to fall freely under the influence of gravity and its velocity is observed, then which of the following statements is correct? (a) The velocity goes on increasing because of the acceleration due to gravity. (b) The velocity goes on increasing until it reaches a maximum. Thereafter, the velocity starts decreasing due to resistive forces. (c) The velocity reaches a maximum and thereafter, the drop falls with a constant velocity. (d) The velocity goes on decreasing and finally the drop becomes stationary.

Two small identical spheres having charges \(+10 \mu \mathrm{C}\) and \(-90 \mu \mathrm{C}\) attract each other with a force of \(F\) newton. If they are kept in contact and then separated by the same distance, the new force between them is (a) \(F / 6\) (b) \(16 \mathrm{~F}\) (c) \(16 \mathrm{~F} / 9\) (d) \(9 F\)

The electric potential at a point \((x, y, z)\) is given by \(V=-x^{2} y-x z^{3}+4 .\) The electric field \(\vec{E}\) at that point is (a) \(\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^{2}+y^{2}\right)+\hat{k}\left(3 x z-y^{2}\right)\) (b) \(\vec{E}=\hat{i z}^{3}+\hat{j} x y z+\hat{k} z^{2}\) (c) \(\vec{E}=\hat{i}\left(2 x y-z^{3}\right)+\hat{j} x y^{2}+\hat{k} 3 z^{2} x\) (d) \(\vec{E}=\hat{i}\left(2 x y-z^{3}\right)+\hat{j} x^{2}+\hat{k} 3 x z^{2}\)

As one penetrates a uniformly charged non-conducting sphere, the electric field strength (a) increases (b) decreases (c) remains constant equal to that at the surface (d) is zero at all points

The mathematical form of Gauss' law is $$ \varepsilon_{o} \oint \vec{E} \cdot d \vec{S}=q $$ In this reference which of the following is correct? (a) \(E\) depends on the charge \(q\) which is enclosed within the Gaussian surface only (b) \(E\) depends on the charge which is inside and outside the Gaussian surface (c) \(E\) does not depend on the magnitude of charge \(q\) (d) All of the above
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