Two electric charges, q1=+20.0 nC and q2=+10.0 nC, are located on the x -axis at x=0 m and x=1.0

step 1 Let us draw the vector diagram. Here is the Q1, and then here is the Q2, right? And the distance

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 Let us draw the vector diagram.

00:03 Here is the q1, and then here is the q2, right? and the distance between q1 and q2 is 1 meter, right? and here is another point with y coordinate 0 .5 meter, right? so let us find out.

00:32 Electric fields at this point due to q1 and q2 well due to q1 so electric field e is in this direction let's say this is e1 and from q2 right electric field is in this direction and this is e2 right now let's resolve e1 in e2 into components well, this is e1, sign of 45 degrees.

01:12 Well, remember that this angle is equal to 45 degrees, since this equals 45, right? then by using geometry, theta is also 45.

01:25 Okay, and the x component of e2 is e2 sine 45, right? so it's e2 -45.

01:40 Okay.

01:43 And further, excuse me, not this angle, rather this angle, it's equal to theta, right? this one.

01:57 This angle is equal to theta.

02:00 This angle is also equal to theta, right? okay, right.

02:10 And the vertical component is like ey is equal to to e1, cosine of 45 plus e2 into cosine of 45.

02:32 And e x, x is equal to e1, sign of 45, right? e x is equal to e1, sine of 45 minus e2, sine of 45 degrees.

03:03 Now let us find out e1 and e2, right? so e1 is equal to k times q divided by r square.

03:16 Well, k is 8 .99 multiplied by 10 to the power 9.

03:21 And q is 20 multiply by 10 to the power minus 9, divided by r, well, it's 0 .7.

03:29 And therefore e1 is equal to 366 .93.

03:38 Now let us find out of e2.

03:40 Well, e2 is equal to k times q divided by r square...

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