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3 years ago

Why do electric field lines explain why like charges repel and opposite charges attract?

1 answer:

5 0

It can be explained as follows: consider the field produced by a positive charge. If we place a positive test charge in this a field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel)

Explanation:

Electric fields are vector fields, and they are represented using field lines.

The field lines give indications on both the magnitude and the direction of the electric field. In fact:

The magnitude of the field can be inferred from the spacing between the lines: the closer the lines are, the stronger the field, while for a weaker field the lines are more spread apart
The direction of the field is given by the direction of the field lines

In particular, by convention the direction of the field lines represent the direction of the force that a positive test charge would feel when immersed in that field: this means that a positive test charge would accelerate in the direction of the field lines, while a negative test charge would accelerate in the direction opposite to the field lines.

This is in agreement with the fact that like charges repel and opposite charges attract. In fact, the lines of the electric field produced by a single-point positive charge point away from the positive charge: if we place a positive test charge in this field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel).

Learn more about electric fields:

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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

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Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

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Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

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and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

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