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
∴
[tex]\rho (r) \ \alpha \ \delta (r -R)[/tex]
[tex]\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)[/tex]
[tex]\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)[/tex]
To find the constant k, we examine the total charge Q which is:
[tex]Q = \int \rho (r) \ dV = \int \sigma \times dA[/tex]
[tex]Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2[/tex]
∴
[tex]\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[/tex]
[tex]\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[/tex]
[tex](2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]
Thus;
[tex]k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2[/tex]
[tex]k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2[/tex]
[tex]k * R^2= \sigma \times R^2[/tex]
[tex]k = R^2 --- (2)[/tex]
Hence, from equation (1), if k = [tex]\sigma[/tex]
[tex]\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}[/tex]
[tex]\mathbf{\rho (r) =0 \ \ at \ \ r<R \ \ or \ \ r>R}[/tex]
To verify the units:
[tex]\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}[/tex]
↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q
[tex]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[/tex]
[tex]Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr[/tex]
[tex]Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr[/tex]
[tex]Q = 4 \pi \sigma *R^2[/tex] since [tex]( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )[/tex]
[tex]\mathbf{Q = 4 \pi R^2 \sigma }[/tex]
