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A particle has charge \(-3.00 \mathrm{nC}\). (a) Find the magnitude and direction of the electric field due to this particle at a point \(0.250 \mathrm{~m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of \(12.0 \mathrm{~N} / \mathrm{C}\) ?

Short Answer

Expert verified
(a) 431.52 N/C upwards; (b) 1.50 m from the charge.

Step by step solution

01

Understanding the problem

We need to calculate the electric field produced by a point charge. We have two parts to this problem: (a) finding the electric field at a specific distance of 0.250 m above the charge, and (b) determining the distance at which the same charge produces an electric field of 12.0 N/C.
02

Formula for Electric Field from a Point Charge

Recall that the electric field (E) due to a point charge can be calculated using the formula: \[ E = \frac{k_e \cdot |q|}{r^2} \] where \( k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is the Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.
03

Substitute and Solve for Part (a)

For part (a), we know \( q = -3.00 \text{ nC} = -3.00 \times 10^{-9} \text{ C} \) and \( r = 0.250 \text{ m} \). Substitute these values into the formula:\[ E = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{(0.250)^2} \]Calculate the value to find the electric field magnitude.
04

Calculation for Part (a)

Continuing the substitution for part (a):\[ E = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{0.0625} \] \[ E \approx 431.52 \text{ N/C} \]The magnitude of the electric field is 431.52 N/C, and its direction is upward, away from the charge (since the charge is negative).
05

Setting Up Part (b)

For part (b), we need to find the distance where the electric field magnitude is 12.0 N/C. Set up the equation:\[ 12.0 = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{r^2} \] and solve for \( r \).
06

Solve for Distance in Part (b)

Solve the equation from Step 5 for \( r \):\[ r^2 = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-9}}{12.0} \]Calculate:\[ r^2 = 2.2475 \]\[ r = \sqrt{2.2475} \approx 1.50 \text{ m} \]Thus, the distance at which the electric field is 12.0 N/C is approximately 1.50 m.

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Key Concepts

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

Coulomb's Constant
In the realm of electromagnetism, Coulomb's Constant (\(k_e\)) is a crucial value that relates electrical force to charge and distance. It is part of the formula that calculates the electric force between two charges, and is crucial for understanding the electric field at a distance from a charge.

Coulomb's Constant is defined as:
  • \(k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\).
  • It is derived from the forces experienced by point charges.
Coulomb's Constant is significant because it is a measure of the force that one point charge exerts on another when they are one meter apart in a vacuum. This value controls how electric fields diminish with distance, governed by the inverse square law.

In calculations, the constant is often paired with the magnitude of a charge and the distance from said charge to determine electric fields or forces present in a system. Its constant nature allows predictions of force under standard conditions.
Point Charge
A point charge refers to an electric charge that is concentrated at a single point in space, having negligible size and volume compared to the distances involved. This ideal model simplifies the calculation of electric fields and forces because the geometry becomes straightforward.

Point charges are useful because:
  • They allow for simplified mathematical models.
  • They provide a clearer conceptual understanding of electric fields.
The electric field generated by a point charge spreads radially outward, not restricted by conductor surfaces or spherical shells. This means we can predict the electric field strength at any point in space using the formula:\[E = \frac{k_e \cdot |q|}{r^2}\],where \(E\) is the magnitude of electric field, \(k_e\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge.
Electric Field Magnitude
The electric field magnitude is a quantitative measure of the strength of an electric field at a specific point. It describes how powerful the field is and how it will influence other charges in its vicinity.

The magnitude can be calculated using:\[E = \frac{k_e \cdot |q|}{r^2}\]In this formula:
  • \(E\) measures how much force a unit positive charge would experience at a point.
  • The magnitude depends inversely on the square of the distance \(r\) from the charge.
This means that the electric field decreases rapidly with increasing distance from the charge. Evaluating the magnitude is crucial in determining how a charge will behave when placed within the field.
Negative Charge
Negative charge implies an excess of electrons resulting in a net negative electrical property. This characteristic affects the nature and direction of the electric field produced by the charge.

Key points about negative charge:
  • Produces electric fields directed toward the charge.
  • Attracts positive charges and repels other negative charges.
The electric field lines around a negative point charge point inward, signifying the direction a positive test charge would move.

Thus, when solving for electric fields, understanding the sign of the charge helps predict field direction, which has effects on interactions with other charged entities. Understanding this concept is fundamental to comprehending electric fields and forces in physics.

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Most popular questions from this chapter

Two very large parallel sheets are \(5.00 \mathrm{~cm}\) apart. Sheet A carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2}\), and sheet \(B\), which is to the right of \(A\), carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2}, .\) Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) \(4.00 \mathrm{~cm}\) to the right of sheet \(A ;\) (b) \(4.00 \mathrm{~cm}\) to the left of sheet \(A ;\) (c) \(4.00 \mathrm{~cm}\) to the right of sheet \(B\).of the net force that these wires exert on it?

A small sphere with mass \(9.00 \mu \mathrm{g}\) and charge \(-4.30 \mu \mathrm{C}\) is moving in a circular orbit around a stationary sphere that has charge \(+7.50 \mu \mathrm{C}\). If the speed of the small sphere is \(5.90 \times 10^{3} \mathrm{~m} / \mathrm{s}\), what is the radius of its orbit? Treat the spheres as point charges and ignore gravity.

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and \(\alpha=30.0^{\circ}\)

A tiny electric dipole \(\left(\mathrm{H}_{2} \mathrm{O}\right.\) modecule) of dipole moment \(\vec{p}\) is placed at a distance \(r\) form an infinitely long wire, with its \(\vec{p}\) normal to the wire in the same plane. If the linear charge density of the wire is \(\lambda\), find the electrostatic force acting on the dipole is equal to

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