Two electric charges, q1=+20.0 nC and q2=+10.0 nC are located on the x -axis at x=0 m and x=1.00

step 1 Okay, so we're told that we have two electric charges. Q1 is positive 20 .0 nanocalooms, and Q2

Two electric charges, q1=+20.0 nC and q2=+10.0 nC are...

Two electric charges, $q_{1}=+20.0 \mathrm{nC}$ and $q_{2}=+10.0 \mathrm{nC}$ are located on the $x$ -axis at $x=0 \mathrm{m}$ and $x=1.00 \mathrm{m}$ respectively. What is the magnitude of the electric field at the point $x=0.50 \mathrm{m}, y=0.50 \mathrm{m} ?$

Key Concepts

Electric Field

The electric field is a vector field that represents the force per unit charge experienced by a test charge placed in space. It is a fundamental concept in electrostatics, indicating both the magnitude and direction of the force that would act on a positive test charge due to the presence of other charges.

Coulomb's Law

Coulomb's Law quantifies the magnitude of the electric force (and thereby the electric field) between two point charges. It states that the force is directly proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. This law forms the basis for calculating the electric field generated by point charges.

Superposition Principle

The Superposition Principle in electrostatics states that the net electric field created by multiple charges is the vector sum of the electric fields produced by each individual charge. This principle allows for the decomposition of complex charge configurations into simpler parts, whereby the contributions from each charge are calculated separately and then added together to determine the overall field.

Vector Addition

Since the electric field is a vector quantity, its calculation in problems involving multiple charges requires vector addition. This involves breaking down each individual electric field into its components and then summing these components along the appropriate axes to obtain the total electric field at a point.

Transcript

00:01 Okay, so we're told that we have two electric charges.

00:02 Q1 is positive 20 .0 nanocalooms, and q2 is positive 10 nanoculums.

00:09 They're located on the x axis at x equals 0 and x equals 1 meter, respectively.

00:16 And we're asked to find the magnitude of the electric field at point x equals 0 .5 meters and y equals 0 .5 meters.

00:26 Okay, so let's go ahead and make a diagram.

00:39 So here's q1 and here's q2.

00:50 It doesn't matter where we put it.

00:53 Okay, and this is one meter.

00:57 This is zero.

01:00 So this will be about five.

01:01 Let me look.

01:13 Okay, so there's my middle and then that's going to be right about here.

01:26 Let's get a new color.

01:30 It didn't work.

01:48 And then we'll have a this, this, have this, go straight from there, this goes straight from there, and this will be the other one.

02:14 Okay, so here we'll call this, we'll call this, okay.

02:40 So, our resulting vector has vertical and horizontal components.

02:47 So, x will be e1 times a sign of theta, minus e2 times the sign of theta, ey will be e1 times a cosine of theta, or i guess not theta, minus e2 times the cosine of our angle.

03:25 Okay.

03:30 And our field strengths at point, the field strength due to charges 1 and 2.

04:02 And then this will be, okay, there we go.

04:19 The electric field due to charges are 1 and 2, and we're going to observe that r1 equals r2.

04:28 And magnitude...

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