Chapter 17: Problem 41
\(\bullet$$\bullet\) A point charge of \(-4.00 \mathrm{nC}\) is at the origin, and a second point charge of \(+6.00 \mathrm{nC}\) is on the \(x\) axis at \(x=0.800 \mathrm{m}\) . Find the magnitude and direction of the electric field at each of the following points on the \(x\) axis: (a) \(x=20.0 \mathrm{cm}\) , (b) \(x=1.20 \mathrm{m},(\mathrm{c}) x=-20.0 \mathrm{cm} .\)
Short Answer
Step by step solution
Identifying Known Values
Applying Electric Field Formula
Calculating Electric Field at \(x=0.20\ \mathrm{m}\)
Calculating Electric Field at \(x=1.20\ \mathrm{m}\)
Calculating Electric Field at \(x=-0.20\ \mathrm{m}\)
Determine Net Electric Field and Direction for Each Point
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Point Charge
In the provided exercise, we have two point charges:
- The first point charge (\(Q_1\)) is \(-4.00 \,\mathrm{nC}\), located at the origin of the coordinate system (\(x = 0\)).
- The second point charge (\(Q_2\)) is \(+6.00 \,\mathrm{nC}\), placed on the \(x\) axis at a distance \(x = 0.800\,\mathrm{m}\).
Understanding how point charges interact is crucial in solving problems concerning electric fields.
Coulomb's Law
Thus, the equation for Coulomb’s Law is given by:
\[ F = k \frac{|Q_1 \cdot Q_2|}{r^2} \]
where \(k\) is Coulomb's constant (\(8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\)).
For electric fields, this law helps determine the field strength (\(E\)) generated by a single point charge at a certain distance.
The electric field from a point charge is computed using:
\[ E = \frac{k |Q|}{r^2} \]
In the exercise, this formula was applied for each point charge to calculate the electric fields at different points on the \(x\)-axis.
Understanding the calculations can aid students in solving similar problems involving the interaction of point charges.
Vector Addition
To find the net electric field at a given point, we must perform vector addition for each field's contribution.
Consider the example at \(x = 0.20\,\mathrm{m}\):
- Calculate the electric field contributed by each point charge using \(E = \frac{k |Q|}{r^2}\).
- Then, consider the direction each field is pointing.
- Since the point charge \(Q_1\) is negative, its field points towards the charge.
- On the other hand, \(Q_2\) being positive causes the field to point away from it.
Mastering vector addition is crucial for determining overall field effects.
Electric Field Direction
An electric field generated by a point charge has a distinct direction:
- From a positive charge, electric fields radiate outwards.
- For a negative charge, these fields point inward, towards the charge.
- The sign of the charge (\(Q\)): negative or positive.
- The position of the charges along the \(x\)-axis, relative to the point of interest.
- For instance, at \(x = 0.20\,\mathrm{m}\), the electric field due to \(Q_1\) (negative) points left towards the origin, while \(Q_2\) (positive) points right.