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In a certain region of space the electric potential is given by \(V = +Ax^2y - Bxy^2,\) where \(A =\) 5.00 \(V/m^3\) and \(B =\) 8.00 \(V/m^3\). Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x =\) 2.00 m, \(y =\) 0.400 m, and \(z = 0\).

Short Answer

Expert verified
The electric field magnitude is approximately 9.85 V/m, directed at about 47.43° from the negative x-axis towards the negative y-axis.

Step by step solution

01

Understanding the electric potential function

The electric potential is given by the function \( V = +Ax^2y - Bxy^2 \), where \( A = 5.00 \, V/m^3 \) and \( B = 8.00 \, V/m^3 \). We need to find the electric field, which is related to the potential by the negative gradient \( \textbf{E} = -abla V \).
02

Calculating the partial derivatives of the potential

The electric field components \( E_x \), \( E_y \), and \( E_z \) are given by:\[ E_x = -\frac{\partial V}{\partial x} \]\[ E_y = -\frac{\partial V}{\partial y} \]\[ E_z = -\frac{\partial V}{\partial z} \]Since \( V \) is not dependent on \( z \), \( E_z = 0 \).
03

Compute the x-component of the electric field

Find \( \frac{\partial V}{\partial x} \):\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(Ax^2y - Bxy^2) = 2Axy - By^2 \]Substitute \( A = 5.00 \, V/m^3 \), \( B = 8.00 \, V/m^3 \), \( x = 2.00 \, m \), and \( y = 0.400 \, m \):\[ \frac{\partial V}{\partial x} = 2(5.00)(2.00)(0.400) - 8.00(0.400)^2 = 8.00 - 1.28 = 6.72 \, V/m \]Thus, \( E_x = -6.72 \, V/m \).
04

Compute the y-component of the electric field

Find \( \frac{\partial V}{\partial y} \):\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(Ax^2y - Bxy^2) = Ax^2 - 2Bxy \]Substitute \( A = 5.00 \, V/m^3 \), \( B = 8.00 \, V/m^3 \), \( x = 2.00 \, m \), and \( y = 0.400 \, m \):\[ \frac{\partial V}{\partial y} = (5.00)(2.00)^2 - 2(8.00)(2.00)(0.400) = 20.00 - 12.80 = 7.20 \, V/m \]Thus, \( E_y = -7.20 \, V/m \).
05

Calculating the magnitude and direction of the electric field

The magnitude of the electric field \( E \) is given by:\[ |\textbf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6.72)^2 + (-7.20)^2 + 0^2} = \sqrt{45.1584 + 51.84} = \sqrt{96.9984} \approx 9.85 \, V/m \]The direction can be found by:\( \theta = \tan^{-1}\left( \frac{E_y}{E_x} \right) = \tan^{-1}\left( \frac{-7.20}{-6.72} \right) \approx \tan^{-1}(1.071) \approx 47.43^\circ \) from the negative x-axis towards the negative y-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 is a measure of the work needed to move a unit charge from a reference point to a specific point within an electric field without any acceleration. It is a scalar quantity that represents the potential energy per unit charge. In our exercise, the electric potential is given by the equation:
\( V = +Ax^2y - Bxy^2 \), where \( A = 5.00 \, V/m^3 \) and \( B = 8.00 \, V/m^3 \).
Electric potential can change based on the position within the field
  • As shown here, the potential is dependent on the coordinates \( x \) and \( y \), and the constants \( A \) and \( B \), which are specific characteristics of this potential field.
  • An understanding of how potential varies is critical because the electric field is derived from the spatial change of this potential.
The concept of electric potential simplifies the mathematics of electromagnetism by reducing vector calculations to simpler scalar operations.
Partial Derivatives
Partial derivatives are used to measure how a function changes as one of the variables changes while holding the other variables constant.
In the context of electric potential, partial derivatives allow us to calculate the components of the electric field from the scalar potential.
For this problem, we need the partial derivatives of \( V \) with respect to \( x \) and \( y \), which correspond to the components \( E_x \) and \( E_y \), respectively:
  • \( \frac{\partial V}{\partial x} = 2Axy - By^2 \)
  • \( \frac{\partial V}{\partial y} = Ax^2 - 2Bxy \)
These derivatives help us understand how swiftly the potential changes in space, which is essential for determining the electric field direction and strength in any specific spot.
Gradient
The gradient is a vector operation that takes a scalar field, like electric potential, and returns a vector field, which in this case is the electric field. This process is critical because it provides information about both the magnitude and the direction of the field.
The electric field \( \bf{E} \) is related to the potential \( V \) through the negative gradient:
  • \( \bf{E} = -abla V \)
The gradient essentially points in the direction of the greatest rate of increase of the scalar field, but we take the negative for electric fields because the field lines point from high to low potential. This vector \( \bf{E} \) tells us how the electric potential is changing spatially, providing the electric field's direction and magnitude at any point in the region.
Vector Magnitude
Vector magnitude is an essential aspect of understanding electric fields as it gives us the intensity of the field. In our problem, once we have the components of the electric field, \( E_x \), \( E_y \), and \( E_z \) (where \( E_z = 0 \)), we calculate the electric field's total magnitude using:
  • \( |\bf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \)
The magnitude tells us how strong the electric field is at the given point.
In the given exercise, we determined:
\( E_x = -6.72 \, V/m \) and \( E_y = -7.20 \, V/m \), leading to a total magnitude:\( \approx 9.85 \, V/m \).
Understanding vector magnitude helps in visualizing just how intense the electric field is at a specific point, influencing how much force a charged particle would experience at that location.

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Most popular questions from this chapter

An object with charge \(q = -6.00 \times 10^{-9}\) C is placed in a region of uniform electric field and is released from rest at point \(A\). After the charge has moved to point \(B\), 0.500 m to the right, it has kinetic energy \(3.00 \times 10^{-7}\) J. (a) If the electric potential at point \(A\) is \(+\)30.0 V, what is the electric potential at point \(B\)? (b) What are the magnitude and direction of the electric field?

A gold nucleus has a radius of 7.3 \(\times 10^{-15}\) m and a charge of \(+79e\). Through what voltage must an alpha particle, with charge \(+2e\), be accelerated so that it has just enough energy to reach a distance of 2.0 \(\times 10^{-14}\) m from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

Two point charges \(q_1 = +\)2.40 nC and \(q_2 = -\)6.50 nC are 0.100 m apart. Point \(A\) is midway between them; point \(B\) is 0.080 m from \(q_1\) and 0.060 m from \(q_2\) (\(\textbf{Fig. E23.19}\)). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B\); (c) the work done by the electric field on a charge of 2.50 nC that travels from point \(B\) to point \(A\).

A helium ion (He\(^{++}\)) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.0 MeV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm = 1 \(\times\) 10\(^{-15}\) m, and \(e\) is the magnitude of the charge of an electron or a proton.) (a) 2\(e\); (b) 11\(e\); (c) 20\(e\); (d) 22\(e\).

For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential \(V\) is zero (take \(V = 0\) infinitely far from the charges) and for which the electric field \(E\) is zero: (a) charges \(+Q\) and \(+2Q\) separated by a distance \(d\), and (b) charges \(-Q\) and \(+2Q\) separated by a distance \(d\). (c) Are both \(V\) and \(E\) zero at the same places? Explain.

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