Class: Electromagnetics. Using the integral how do you get from one part to the other From => Q = C₁E { [ { [ 4 π S³ (²) - (₂² @ ²2² ] = r² sin odrded D(7)-(2²cr2rshow work16. Total→ To => Q = C₁ E₂ [4π-4πC₂²² [e-<₂²-rdr]Sult2O

The value of Q is given as Q=C1ε0∫0∞∫0π∫02π4πδ3(r→)-C22e-C2rrr2sinθdrdθdϕ This integration can be…

Answered: From = Q = C₁E { [4π 8³(P)-(₂²=(²²) ²…

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Class: Electromagnetics. Using the integral how do you get from one part to the other

From => Q = C₁E { [ { [ 4 π S³ (²) - (₂² @ ²2² ] = r² sin odrded D
(7)-(2²
cr
2
r
show work
16. Total
→ To => Q = C₁ E₂ [4π-4πC₂²² [e-<₂²-rdr]
Sult
2
O

Expert Solution

Step 1

The value of Q is given as

$Q = C_{1} ε_{0} \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} [4 {πδ}^{3} (\vec{r}) - C_{2}^{2} \frac{e^{- C_{2} r}}{r}] r^{2} sinθdrdθdϕ$

This integration can be broken into two parts

$Q = C_{1} ε_{0} \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} 4 {πδ}^{3} (\vec{r}) r^{2} sinθdrdθdϕ - C_{1} ε_{0} \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} C_{2}^{2} \frac{e^{- C_{2} r}}{r} r^{2} sinθdrdθdϕ$

Take the constant out of integration

$Q = C_{1} ε_{0} 4 π \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} δ^{3} (\vec{r}) r^{2} sinθdrdθdϕ - C_{1} ε_{0} C_{2}^{2} \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} \frac{e^{- C_{2} r}}{r} r^{2} sinθdrdθdϕ$

$Q = C_{1} ε_{0} 4 π \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} δ^{3} (\vec{r}) r^{2} sinθdrdθdϕ - C_{1} ε_{0} C_{2}^{2} \int_{0}^{\infty} \int_{0}^{π} \int_{0}^{2 π} e^{- C_{2} r} rsinθdrdθdϕ$ ............................ (1)

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From = Q = C₁E { [4π 8³(P)-(₂²=(²²) ² sin earded b ( w work 1b. Total \T. IS THE LIN ²0-4₂² ordr