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3 years ago

12

Particle a has twice the charge of nearby particle

1 answer:

bulgar [2K]3 years ago

8 0

The force between charged particles is calculated through the Coulomb's law,
F = kQ₁Q₂/d²
From the equation, it can be deduced that the force is linearly directly proportional to the charge. This means that when the charge is doubled, the force is likewise doubled. Therefore, the compared to force on b, the force on a is twice.

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3 years ago

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

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2 years ago