Question

A thin nonconducting spherical shell of radius R1 carries a total charge q1 that is uniformly distributed on its surface. A second, larger thin non-conducting spherical shell of radius R2 that is concentric with the first carries a charge q2 that is uniformly distributed on its surface.

Required:
a. Use Gauss's law to find the electric field in the regions r < R1, R1 < r < R2, and r > R2.
b. What should the ratio of the charges q1/q2 and their relative signs be for the electric field to be zero for r > R2?

Answer 1

Answer:

Explanation:

a )

in the regions r < R₁

charge q inside sphere of radius R₁ = 0

Applying gauss's law for electric field E at distance r <R₁

electric flux through Gaussian surface of radius r = 4π r² E

4π r² E = q / ε₀ = 0 / ε₀

E = 0

Applying gauss's law for electric field E at distance R₁ < r < R₂ .

charge q inside sphere of radius R₁ = q₁

Applying gauss's law for electric field E at distance R₁ < r < R₂

electric flux through Gaussian  surface of radius r = 4π r² E

4π r² E = q₁ / ε₀

E = q₁ / 4πε₀

in the regions r> R₂

charge q inside sphere of radius R₂ = (q₁ + q₂)

Applying gauss's law for electric field E at distance r > R₂

electric flux through Gaussian surface of radius r = 4π r² E

4π r² E = (q₁ + q₂) / ε₀

E =  (q₁ + q₂) /4π ε₀

b )

For electric flux to be zero at r > R₂

(q₁ + q₂) /4π ε₀  = 0

q₁ + q₂ = 0

q₁ / q₂ = - 1 .

Answer 2

(a) The value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be $\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}$

(b)The ratio of the charges q1/q2 will be -1.

What is gauss law?

The charge contained divided by the permittivity equals the total electric flux out of a closed surface, according to Gauss Law.

The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane parallel to the field.

Let's analyse regions r < R₁

charge q inside sphere will be zero.

According to gauss's law for electric field E at distance r <R₁

electric flux for the gaussian surface  = 4π r² E

$4\pi r^2E=\frac{Q}{\epsilon_0} =0$

By the applications of  gauss's law for electric field E at distance R₁ < r < R₂ .

q is charge inside sphere having radius R₁ = q₁

By the applications of gauss's law for electric field E at distance R₁ < r < R₂

electric flux obtained due to Gaussian surface of radius r = 4π r² E

$4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1}{\epsilon_0}$

For the regions r> R₂

q is the charge inside sphere of radius hyaving  R₂ = (q₁ + q₂)

By the application of gauss's law for electric field E which is at distance r > R₂

Electric flux through Gaussian surface of radius r = 4π r² E

$4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1+q_2}{\epsilon_0}$

Hence the value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be $\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}$.

(b) To finding the ratio of charges the value of  electric flux must be zero at r > R₂

$\frac{(q_1+ q_2) }{4\pi\epsilon_0} =0\\\\q_1 + q_2= 0\\\\\frac{q_1}{q_2} =-1$

Hence the ratio of the charges q1/q2 will be -1.

To learn more about the gauss law refer to the link;

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