Two charges are placed on the x axis. One of the charges (q1=+8.5 μC) is at x1=+3.0 cm and the o

Two charges are placed on the x axis. One of the charges...

Two charges are placed on the $x$ axis. One of the charges $\left(q_{1}=+8.5 \mu \mathrm{C}\right)$ is at $x_{1}=+3.0 \mathrm{cm}$ and the other $\left(q_{2}=-21 \mu \mathrm{C}\right)$ is at $x_{1}=+9.0 \mathrm{cm} .$ Find the net clectric ficld (magnitude and dircction) at $(\mathrm{a}) x=0 \mathrm{cm}$ and $(\mathrm{b}) x=+6.0 \mathrm{cm} .$

Key Concepts

Superposition Principle

The superposition principle is fundamental when dealing with multiple charges, as it allows for the net electric field at any point to be determined by vectorially adding the individual fields produced by each charge. This linear combination of fields is crucial for solving complex electrostatic problems.

Vector Nature and Direction of Fields

Since the electric field is a vector, its calculation involves careful consideration of both magnitude and direction. The sign of the source charges influences the field's direction: fields emanate from positive charges while they converge towards negative charges, and this directional information is pivotal when combining fields from multiple sources.

Electric Field

An electric field is a vector quantity that represents the force experienced per unit charge at any point in space due to the presence of charges. It describes both the magnitude and direction of the force a positive test charge would experience and forms the basis for understanding interactions in electrostatics.

Coulomb's Law

Coulomb's Law quantifies the electric field created by a point charge, stating that the field's magnitude is directly proportional to the source charge's magnitude and inversely proportional to the square of the distance from the charge. This law introduces the constant of proportionality, ensuring that the calculations are physically accurate.

Transcript

00:01 For this question, for the electric field, i just want to use the equation, electric field, k times the charge over the distance.

00:18 So for the first one, just simply we can plug it in.

00:25 K -conson 8 .99 times 10 to the 9.

00:29 The charge, the first charge, is 8 .5 times 10 to the negative 6 cools.

00:34 The distance is x1 minus x squared which is 0 .03 squared meter squared so you can just rewrite that out and when you plug that in you get 8 .5 times 10 to the 7 you can use per kool.

01:13 Now since the charge is positive it acts towards the left or like the negative x -rection.

01:21 Similarly the magnet to electric field at x equals 0 from the second charge charge, just use 9, i think things easier.

01:35 The second charge is 21 times 10 to negative 6.

01:40 Since it's the magnitude of the charge and not the direction, it doesn't really matter about putting in the negative sign.

01:51 And for this, the distance is 0 .09 meters.

01:57 We're using a meter instead of centimeters since it's the essay units easier to use.

02:04 And that gives you 2 .3 times 10 to the 7 units.

02:07 P .m.

02:09 So as the charge is negative, it acts along the positive x direction.

02:14 So it acts in the opposite direction of the first one we found, the field.

02:18 So the net magnitude of the field at x equals 0, just defined by subtracting.

02:28 And you get 6 .2 times 10 to the 7, newtons per kuhum.

02:34 And this, the total magnitude, the final magnitude, the net direction, is towards the left and the negative x...

Please give Ace some feedback