Q. 59 A sphere of radius R聽has total ... [FREE SOLUTION]

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Chapter 24: Q. 59 (page 686)

A sphere of radius $R$ has total charge $Q$ . The volume charge Calc density role="math" localid="1648722354966" $(\frac{C}{m^{3}})$ within the sphere is $蚁 (r) = \frac{C}{r^{2}}$ , where $C$ is a constant to be determined.
a. The charge within a small volume $d V$ is $d q = 蚁 d V$ . The integral of $蚁 d V$ over the entire volume of the sphere is the total charge $Q$ . Use this fact to determine the constant $C$ in terms of $Q$ and $R$ .
Hint: Let $d V$ be a spherical shell of radius $r$ and thickness $d r$ . What is the volume of such a shell?
b. Use Gauss's law to find an expression for the electric field strength $E$ inside the sphere, , $r 鈮� R$ in terms of $Q$ and $R$ .
c. Does your expression have the expected value at the surface, $r = R$ ? Explain.

Short Answer

Expert verified

Part $(a)$

$(a)$ The value of constant is,role="math" localid="1648723245867" $C = \frac{Q}{4 R 蟺}$ .

Part $(b)$

$(b)$ The expression of electric field strength is, $E = \frac{1}{4 蟺系_{0}} \frac{Q}{R r}$ .

Part $(c)$

$(c)$ The expected value at the surface, $E = \frac{Q}{4 蟺系_{0} R^{2}}$ .

Step by step solution

Step: 1 Finding the constant: (part a)

The charge within a small volume $d V$ of expression is

$\begin{array}{l} \begin{array}{r} d q = 蚁 d V \end{array} \\ d V = 4 r^{2} 蟺 d r \\ d q = \frac{C}{r^{2}} 鈰� 4 r^{2} 蟺 d r \\ Q = 4 蟺 C {鈭�}_{0}^{R} 鈥� d r \\ Q = 4 蟺 C [r]_{0}^{R} \\ Q = 4 蟺 C R \\ C = \frac{Q}{4 R 蟺} . \end{array}$

Step: 2 Expression of electric field strength: (part b)

For $r 鈮� R$ by using Gauss's law finding the electric field.

Selecting the spherical gaussian surface.

localid="1648724918692" $\begin{array}{l} 鈭� \overset{鈫�}{E} 鈰� d \overset{鈫�}{A} = \frac{Q_{i n}}{系_{0}} \\ 鈭� \overset{鈫�}{E} 鈰� d \overset{鈫�}{A} = E A \\ E A = \frac{Q_{i n}}{系_{0}} \\ E = \frac{Q_{i n}}{A 系_{0}} \\ A = 4 r^{2} 蟺 \\ Q_{i n} = 鈭� 蚁 d V \\ Q_{i n} = 4 蟺 C {鈭�}_{0}^{r} 鈥� d r \end{array}$

Step: 3 Expression of electric field strength: (part b)

To get charged inside surface by integrating from $0$ to $r$

$\begin{array}{l} Q_{i n} = 4 蟺 C [r]_{0}^{r} \\ Q_{i n} = 4 蟺 C r \\ Q_{i n} = 4 蟺 \frac{Q}{4 R 蟺} r \\ Q_{i n} = \frac{Q r}{R} \\ E = \frac{Q r}{R} 鈰� \frac{1}{4 r^{2} 蟺系_{0}} \\ E = \frac{1}{4 蟺系_{0}} \frac{Q}{R r} . \end{array}$