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A. What is the electric field strength at r = 19.0 cm?

B. What is the electric flux through a 38.0-cm-diameter spherical surface that is concentric with the charge distribution?
C. How much charge is inside this 38.0-cm-diameter spherical surface?

Question is in the attachment picture!

A. What is the electric field strength at r = 19.0 cm? B. What is the electric flux-example-1
User Arathunku
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1 Answer

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Answer:

Step-by-step explanation:

The electric field produced by a spherically symmetrical charge distribution is given by:

E = kQr / r^3

where k is Coulomb's constant (k = 1 / (4πε0)), Q is the total charge, and r is the distance from the center of the distribution.

A. What is the electric field strength at r = 19.0 cm?

To find the electric field strength at r = 19.0 cm, we need to substitute r = 0.19 m into the given expression for E:

E = (4000(0.19)^2)(0.19)^N/C

We do not have a value for N, so we cannot calculate E. Please provide the value of N.

B. What is the electric flux through a 38.0-cm-diameter spherical surface that is concentric with the charge distribution?

The electric flux through a spherical surface of radius r is given by:

Φ = E * 4πr^2

where E is the electric field strength at the surface of the sphere, and r is the radius of the sphere.

In this case, the diameter of the spherical surface is 38.0 cm, so the radius is r = 0.19 m. We do not have a value for N, so we cannot calculate E. Please provide the value of N.

C. How much charge is inside this 38.0-cm-diameter spherical surface?

The charge inside a spherical surface is given by:

Q = (4/3)πr^3 * ρ

where r is the radius of the sphere, ρ is the charge density, and the factor of (4/3)πr^3 is the volume of the sphere.

In this case, the diameter of the spherical surface is 38.0 cm, so the radius is r = 0.19 m. We do not have a value for ρ, so we cannot calculate Q. Please provide the value of ρ.

User Waterlink
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