Three charges (qA, qB., and qC ) are placed at the vertices...

Three charges $\left(q_{A}, q_{B}\right.$, and $q_{C}$ ) are placed at the vertices of the equilateral triangle that has sides of length $s$ in Figure 16-31. Derive expressions for the electric field at (a) $X$ (at the center of the triangle), (b) $Y$ (at the midpoint of the side between $q_{\mathrm{B}}$ and $\left.q_{\mathrm{C}}\right)$, and (c) $Z$ (at the midpoint of the side between $q_{A}$ and $q_{C}$ ). (d) Now use the following numerical values and calculate the electric field at those same points: $\mathrm{s}=10.0 \mathrm{~cm},^{s}=10.0 \mathrm{~cm}$, $\mathrm{qA}=+20.0 \mathrm{nC}, q_{\mathrm{A}}=+20.0 \mathrm{nC}, \mathrm{qB}=-8.00 \mathrm{nC}, q_{\mathrm{B}}=-8.00 \mathrm{nC}$, and $\mathrm{qC}=-10.0 \mathrm{nC}$ $q_{\mathrm{C}}=-10.0 \mathrm{nC}_{\underline{ }}$ Example $16-7$

Three charges $\left(q_{A}, q_{B}\right.$, and $q_{C}$ ) are placed at the vertices of the equilateral triangle that has sides of length $s$ in Figure 16-31. Derive expressions for the electric field at (a) $X$ (at the center of the triangle), (b) $Y$ (at the midpoint of the side between $q_{\mathrm{B}}$ and $\left.q_{\mathrm{C}}\right)$, and (c) $Z$ (at the midpoint of the side between $q_{A}$ and $q_{C}$ ). (d) Now use the following numerical values and calculate the electric field at those same points: $\mathrm{s}=10.0 \mathrm{~cm},^{s}=10.0 \mathrm{~cm}$, $\mathrm{qA}=+20.0 \mathrm{nC}, q_{\mathrm{A}}=+20.0 \mathrm{nC}, \mathrm{qB}=-8.00 \mathrm{nC}, q_{\mathrm{B}}=-8.00 \mathrm{nC}$, and $\mathrm{qC}=-10.0 \mathrm{nC}$
$q_{\mathrm{C}}=-10.0 \mathrm{nC}_{\underline{ }}$ Example $16-7$

Key Concepts

Coulomb's Law

Coulomb's law gives the magnitude and direction of the force between two point charges. It is foundational in determining how charges interact in space, stating that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, with the electric field arising from a point charge defined as this force divided by a test charge.

Electric Field

The electric field is a vector field that represents the force per unit charge experienced by a positive test charge at any point in space. It is used to describe how charges influence each other nonlocally and is typically calculated by dividing the force acting on a test charge by the magnitude of that charge.

Principle of Superposition

The superposition principle states that the total electric field produced by multiple point charges is the vector sum of the electric fields produced by each individual charge. This principle is crucial when dealing with systems of multiple charges, as it allows the calculation of the net field by considering each charge's contribution independently.

Symmetry in Electrostatics

Symmetry considerations in electrostatics are used to simplify the analysis of electric fields in configurations where the geometric arrangement of charges exhibits symmetry. For example, in an equilateral triangle, the symmetry helps in predicting which components of the electric field will cancel out at certain points, thereby reducing the complexity of the vector summations.

Geometric Analysis in Charge Configurations

Geometric analysis involves using the spatial arrangement of charges, such as distances, angles, and midpoints, to formulate expressions for the electric field. Understanding the geometric properties of arrangements like an equilateral triangle enables the derivation of distances and directional components necessary for applying Coulomb's law and the superposition principle.

Transcript

00:01 For this problem on the topic of electrostatics, we have three charges, a, b and c, placed at the vertices of an equilateral triangle that has a side length of s as shown in the figure.

00:11 We want to derive expressions for the electric field at point x, y, and then point z, and then use the given numerical values to calculate the electric field at the same points.

00:24 Now, firstly, to find the x component at x, ex, is equal to the contribution from each charge, e, e, a x plus the field due to b e bx plus that due to c e cx and this is equal to zero plus k q b divided by s over the square root of three all squared multiplied by the cosine of 30 degrees minus k q c divided by over the square root of 3 all squared times the cosine of 30 degrees and so this becomes 3 k k k b over s squared times the square root of 3 over 2 minus 3 k q c over s squared times the square root of 3 over 2 minus over 2, and we can write this as 3k times the square root of 3 over 2s squared into qb minus qc.

02:00 Next we can find the y component at x, which is ey, and that again is e, a, y, plus e, b, y, plus e, c, minus k k k k a over s over the square root of three all squared plus k q b over s over root 3 all squared times the sign of 30 degrees plus k q c c c divided by by s over the square root of 3 all squared multiplied by the sign of 30 degrees.

03:05 So if we calculate this and simplify, we eventually get 3k over s squared into minus qa plus qb over 2 plus qc over 2.

03:31 Now for part b, the x component at y, ex is equal to eax plus ebx plus e cx, like before, which is 0 plus k qb over s over 2 all squared minus kqc divided by s over 2 all squared.

04:05 And this, after solving and simplifying, becomes 4k over s squared into qb minus qc.

04:18 Next you want to find the y component at y.

04:24 Ey is equal to e -a -y plus e -b -y plus e -c -y, which is minus k -qa over s times the square root of 3 over 2.

04:41 O squared plus 0 plus 0.

04:46 And so we are left with minus 4 kqa divided by 3s squared.

05:01 Now for part c we want to find the x and y components at z.

05:05 And again, x is x a, x plus ebx plus ecx.

05:10 And so the x component at z is kqa.

05:17 Over s over 2 all squared, multiplied by the cosine of 60 degrees, plus kqb over s times the square root of 3 over 2 all squared, multiplied by the cosine of 30 degrees, minus k qc divided by s over 2 all squared, all times the cosine of 60 degrees...

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