CALC Positive charge +Q is distributed uniformly along the +x -axis from x=0 to x=a . Negative c

CALC Positive charge +Q is distributed uniformly along the...

Key Concepts

Electric Dipole Field

An electric dipole consists of two equal and opposite charges separated by a small distance. The field of a dipole in the far-field region falls off as the inverse cube of the distance from the dipole. This behavior is significant when analyzing systems where the leading-order term in the multipole expansion is the dipole moment, as it determines how the electric force decreases with distance in such configurations.

Superposition Principle

The superposition principle in electrostatics states that the net force or electric field produced by multiple charge elements is the vector sum of the individual forces or fields from each charge element. This principle allows one to compute the influence of complex charge configurations by breaking them down into simpler parts and adding the effects from each, making it an essential tool in analyzing problems involving multiple charges.

Multipole Expansion

Multipole expansion is an approximation method used to describe the potential and field due to a spatial charge distribution at points far from the distribution. In such cases, the distribution is expressed in terms of simpler components such as monopole, dipole, quadrupole moments, etc. For a neutral configuration (net charge zero), the dipole term dominates the far-field behavior, leading to a field that decays more rapidly with distance, such as in the proportionality to the inverse cube of the distance.

Continuous Charge Distribution

When charges are spread out over a region rather than concentrated at points, the concept of a continuous charge distribution is used. In these cases, the overall force or electric field is calculated by integrating the contributions from each differential element of charge. This approach is critical in obtaining accurate results for fields and forces in systems where the charge is distributed along lines, surfaces, or volumes.

Coulomb's Law

Coulomb's Law is the fundamental principle governing the interaction between electric charges. It states that the magnitude of the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In problems involving extended charge distributions, Coulomb's Law is applied to infinitesimal charge elements, and the total force is obtained by integrating these contributions over the entire distribution.

Transcript

00:01 For problem 95, we have a positive line of charge from 0 to a and a negative line of charge from 0 to negative a.

00:10 And we have a small q on the positive axis.

00:14 Positive y axis, we have a queue.

00:18 And we need to find the electric force exerted by the line of charge on this small q.

00:27 And we need to show that this electric force is proportional to.

00:31 Y raise to negative 3 for y is much greater than a so first let's draw the line of force exerted by the line of charges so let's call this very small line as the x let's call the distance of this point to the y axis as x and this one as our y this would be the distance d which is the square of x squared plus y squared this angle as theta and we have a repulsion force let's call it as f1 an attraction force let's call this f2 so our f1 has a y n x component also same with our f2 so we know that our fy is equal to zero because the component of f1y and f2y is equal in magnitude.

01:46 So let's solve for the net horizontal force.

01:54 So we have negative to f1 x is equal to negative to f2x.

02:02 So we only need to solve either one of this.

02:06 So let's first define our f1.

02:09 So we have k q qd divided by the distance d squared.

02:16 And if we take the derivative of this, we would have the f1 as equals to kq, dq, divided by the distance, which is x squared plus k squared, and we can substitute this dq as lambda dx.

02:38 And for the f1x, it is equal to the f1 multiplied by the cosine of theta.

02:47 But the cosine of theta is equal to x all over the square of x squared plus y squared.

02:57 So let's substitute our values here.

03:00 So we would have kq lambda d x divided by x squared plus y squared multiplied by x divided by the square of x squared plus y squared.

03:17 And simplifying this, we would get kq lambda.

03:22 X d x divided by x squared plus y squared raise to three halves we would have f1 x is equals to the integral of kq lambda x d x divided by x squared plus y squared raise to three hubs and if we integrate this we would have k q lambda integral of x d x divided by x squared plus y squared these to three halves from zero to a so evaluating this we would have we would have negative one all over the square of x squared plus y squared from zero to a and then we would have negative one all over y squared plus a squared plus one plus one all over y and if we multiply this by two we would have our net force fx so we have negative two because our net force horizontal is to the left so we have the negative here we have negative two kq lambda multiplied by negative one all over the square of y squared plus a squared plus 1 all over y so we can substitute lambda and we can factor out the negative here so we would have 2 k q q all over a multiplied by 1 all over the square root of y squared plus a squared minus 1 all over y in the negative x direction so this is our net force x and show that our net force is proportional to y -raise 2 negative 3 let's take this fraction and approximate it so we have y squared plus a squared is to negative 1 half multiplied by the square root of y -rease 2 negative 2 divided by the square of y -raise 2 negative 2 so we have 1 over y multiplied by 1 plus a squared divided by y squared is to negative 1 half.

06:38 So we can approximate this 1 plus a squared over y squared.

06:43 So we have 1 over y plus negative 1 half times a squared y squared by y plus negative 1 half negative 1 half minus 1 multiplied by a raise to the 2.

07:03 4 divided by y raise 2 the 4 divided by 2 factorial times y plus and so on and this so this is very small so we can ignore this it it is negligible and our approximation will give us 1 over y minus a squared divided by 2 pi cube so let's plug it in our equation again so we would have the net force is equal to 2 kqq all over a multiplied by 1 over y minus a squared divided by 2 y cube minus 1 over y.

07:56 We can subtract this and we can further simplify.

08:03 So our equation give us kq cubed divided by y cube...

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