Is it possible for an object to have a charge of 4.8x10-9 C? Why or why not?

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

2 years ago

Why do the graphs differ?

melisa1 [442]

Well first graph represents rectangular hyperbola

vu = c^2 ( c is constant)

AS 1/v + 1/u = 1/f

Take1/ f to be constant c

1/v = c - 1/u

it is of the form y = - x + k

Slope = -1 having intercept k as shown in fig 2

3 0

3 years ago

When you are riding in a car then the driver suddenly steps on the gas pedal, why do you feel a jolt and push back?

inessss [21]

Answer:

due to acceleration lol

4 0

2 years ago

Read 2 more answers

The dark bands on the wall are from ______________ interference. and i will give brainlyest

Fittoniya [83]

Light can be seen as an electromagnetic wave.
What happens when two waves, with the same frequency, superpose is called interference.

If at a certain point two waves arrive both with a crest, we have constructive interference and the amplitudes sum up, reaching the maximum value, resulting in bright spots.

If at a certain point one of the waves arrives with a crest and the other wave arrives with a trough, we have destructive interference, and the two amplitudes cancel out, resulting in dark spots.

Therefore, t<span>he dark bands on the wall are from destructive interference.</span>

4 0

3 years ago

Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm.

Elodia [21]

Answer:

Answer

Correct option is

A

5×10

−6

tesla

I=5A

x=0.2m

Magnetic field at a distance 0.2 m away from the wire.

B=

2πx

μ

0

I

=

2π×0.2

4π×10

−7

×5

=10×5×10

−7

=5×10

−6

tesla

3 0

2 years ago