Ask question

Ask question

All categories

3 years ago

11

What is the electric field 3.3 m from the center of the terminal of a Van de Graaff with a 7.20 mC charge, noting that the field

is equivalent to that of a point charge at the center of the terminal?

1 answer:

Snowcat [4.5K]3 years ago

6 0

Answer:

Electric Field is $5.943801*10^6 N/C$

Explanation:

Electric Field:

It originates from positive charge and ends at negative charge.

General Formula for electric Field:

$E=\frac{kq}{r^2}$

where:

k is the Coulomb Constant

q is the charge

r is the distance

Given:

q=7.20 mC

r=3.3 meters

k= $8.99*10^9 N.m^2/C^2$

Find:

Electric Field=?

Solution:

$E=\frac{kq}{r^2}$

$E=\frac{(8.99*10^9)(7.20*10^{-3})}{3.3^2}\\E=5943801.653 N/C\\E=5.943801653*10^6 N/C$

Electric Field is $5.943801*10^6 N/C$

You might be interested in

The unit of work is called derived unit.Why

Artemon [7]

Answer:

the unit of work is derived unit because joule is defined the work done by the force aftab 1 newton causing the displacement of one metre something newton metre(n-m) is also used to measuring work.

6 0

3 years ago

Why are triangles important when making structures

Alex_Xolod [135]

So when making structures you have to have squares rectangles right and a triangle and another triangle equals a square so thats all there is to it

4 0

3 years ago

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

3 years ago

If the same experiment is repeated in different parts of the world by different scientists,

Misha Larkins [42]

The answer is D) The outcome of the experiment will be non observable

3 0

3 years ago

Read 2 more answers

A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.

Andreyy89

Answer:

$I_{total}=220.64 kg*m^{2}$

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

Inertia disk

$I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}$

Inertia person

$I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}$

Inertia dog

$I_{d}=\frac{1}{2}*m_{d}*(r_{d})^{2}$

The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

$I_{total}=I_{disk}+I_{p}+I_{d}$

$I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}$

$I_{total}=220.64 kg*m^{2}$

5 0

3 years ago