Let's solve each part step by step: --- ## **(a) Electric Field of a Charged Disk at a Point on Its Axis** **Given:** - Disk radius: \( a \) - Surface charge density: \( \rho_s \) - Disk lies in the \( yz \)-plane at \( x = x_0 \) - The central axis is along the \( x \)-axis. - Find \( \vec{E} \) at \( (x, 0, 0) \) (i.e., on the \( x \)-axis). --- ### **Step 1: Geometry and Charge Element** - Every point on the disk has coordinates \( (x_0, y', z') \). - Let’s use cylindrical coordinates: \( y' = r' \cos \theta, \ z' = r' \sin \theta \), with \( 0 \leq r' \leq a \), \( 0 \leq \theta < 2\pi \). --- ### **Step 2: Vector from Source to Field Point** Field point: \( (x, 0, 0) \) Source point: \( (x_0, y', z') \) Displacement vector: \[ \vec{R} = (x - x_0)\hat{x} - y' \hat{y} - z' \hat{z} \] Magnitude: \[ |\vec{R}| = \sqrt{(x-x_0)^2 + y'^2 + z'^2} = \sqrt{(x-x_0)^2 + r'^2} \] --- ### **Step 3: Differential Contribution to Field** The field due to an infinitesimal surface charge \( dQ = \rho_s \, dS \): \[ d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{dQ \, \vec{R}}{|\vec{R}|^3} \] But due to symmetry, only the \( x \)-component survives at \( (x,0,0) \). \[ dE_x = \frac{1}{4\pi\epsilon_0} \frac{\rho_s \, dS \, (x - x_0)}{[(x-x_0)^2 + r'^2]^{3/2}} \] Where \( dS = r' dr' d\theta \). --- ### **Step 4: Integrate Over the Disk** Integrate over \( r' \) and \( \theta \): \[ E_x = \int_{0}^{2\pi} \int_{0}^{a} \frac{1}{4\pi\epsilon_0} \frac{\rho_s r' dr' d\theta (x - x_0)}{[(x-x_0)^2 + r'^2]^{3/2}} \] \[ E_x = \frac{\rho_s (x - x_0)}{4\pi\epsilon_0} \int_{0}^{2\pi} d\theta \int_{0}^{a} \frac{r' dr'}{[(x-x_0)^2 + r'^2]^{3/2}} \] \[ E_x = \frac{\rho_s (x - x_0)}{2\epsilon_0} \int_{0}^{a} \frac{r' dr'}{[(x-x_0)^2 + r'^2]^{3/2}} \] --- ### **Step 5: Solve the Integral** Let \( u = (x-x_0)^2 + r'^2 \), \( du = 2r' dr' \), so \( r' dr' = \frac{du}{2} \): For \( r' = 0 \), \( u_1 = (x-x_0)^2 \) For \( r' = a \), \( u_2 = (x-x_0)^2 + a^2 \) \[ \int_0^a \frac{r' dr'}{[(x-x_0)^2 + r'^2]^{3/2}} = \int_{u_1}^{u_2} \frac{du}{2u^{3/2}} \] \[ = \frac{1}{2} \int_{u_1}^{u_2} u^{-3/2} du = \frac{1}{2} \left[ -2u^{-1/2} \right]_{u_1}^{u_2} \] \[ = -u^{-1/2} \Big|_{u_1}^{u_2} = \left[ \frac{1}{\sqrt{u_1}} - \frac{1}{\sqrt{u_2}} \right] \] \[ = \frac{1}{|x-x_0|} - \frac{1}{\sqrt{(x-x_0)^2 + a^2}} \] --- ### **Step 6: Final Expression** \[ E_x = \frac{\rho_s (x - x_0)}{2\epsilon_0} \left[ \frac{1}{|x-x_0|} - \frac{1}{\sqrt{(x-x_0)^2 + a^2}} \right] \] If \( x > x_0 \), \( |x-x_0| = x-x_0 \): \[ \boxed{ E_x(x,0,0) = \frac{\rho_s}{2\epsilon_0} \left[1 - \frac{x-x_0}{\sqrt{(x-x_0)^2 + a^2}}\right] } \] This is the electric field **along the axis** of a uniformly charged disk. --- ## **(b) Electric Field on the Axis of a Charged Cylinder (from result in a)** - Cylinder: Radius \( a \), height \( L \), uniform volume charge density \( \rho_v \) - Axis along \( x \), cylinder occupies \( x_0 \leq x_1 \leq x_0+L \) - Find \( E_x \) at \( (x,0,0) \) --- ### **Step 1: Consider the Cylinder as Stacked Disks** Each infinitesimal disk at \( x_1 \) has thickness \( dx_1 \) and surface charge density \( \rho_s = \rho_v dx_1 \). The field at \( (x,0,0) \) from each such disk (from part a) is: \[ dE_x = \frac{\rho_s}{2\epsilon_0} \left[1 - \frac{x-x_1}{\sqrt{(x-x_1)^2 + a^2}}\right] \] But \( \rho_s = \rho_v dx_1 \): \[ dE_x = \frac{\rho_v \, dx_1}{2\epsilon_0} \left[1 - \frac{x-x_1}{\sqrt{(x-x_1)^2 + a^2}}\right] \] --- ### **Step 2: Integrate Over the Cylinder** Integrate \( x_1 \) from \( x_0 \) to \( x_0 + L \): \[ E_x = \frac{\rho_v}{2\epsilon_0} \int_{x_0}^{x_0+L} \left[1 - \frac{x-x_1}{\sqrt{(x-x_1)^2 + a^2}}\right] dx_1 \] \[ = \frac{\rho_v}{2\epsilon_0} \left\{ \int_{x_0}^{x_0+L} 1 \, dx_1 - \int_{x_0}^{x_0+L} \frac{x-x_1}{\sqrt{(x-x_1)^2 + a^2}} dx_1 \right\} \] \[ = \frac{\rho_v}{2\epsilon_0} \left\{ L - \int_{x_0}^{x_0+L} \frac{x-x_1}{\sqrt{(x-x_1)^2 + a^2}} dx_1 \right\} \] --- ### **Step 3: Solve the Remaining Integral** Let \( u = x-x_1 \), so \( du = -dx_1 \): When \( x_1 = x_0 \), \( u = x - x_0 \) When \( x_1 = x_0 + L \), \( u = x - x_0 - L \) So, \[ \int_{x_0}^{x_0+L} \frac{x-x_1}{\sqrt{(x-x_1)^2 + a^2}} dx_1 = \int_{u_1}^{u_2} \frac{u}{\sqrt{u^2 + a^2}} (-du) = -\int_{u_1}^{u_2} \frac{u}{\sqrt{u^2 + a^2}} du \] \[ = - \left[ \frac{1}{2} \sqrt{u^2 + a^2} \right]_{u_1}^{u_2} \] Where \( \int u / \sqrt{u^2 + a^2} du = \frac{1}{2} \sqrt{u^2 + a^2} \). So, \[ = - \left( \frac{1}{2} \sqrt{u_2^2 + a^2} - \frac{1}{2} \sqrt{u_1^2 + a^2} \right) = \frac{1}{2} \left[ \sqrt{(x-x_0)^2 + a^2} - \sqrt{(x-x_0-L)^2 + a^2} \right] \] --- ### **Step 4: Final Expression** \[ E_x = \frac{\rho_v}{2\epsilon_0} \left\{ L - \frac{1}{2} \left[ \sqrt{(x-x_0)^2 + a^2} - \sqrt{(x-x_0-L)^2 + a^2} \right] \right\} \] Or, boxed: \[ \boxed{ E_x(x,0,0) = \frac{\rho_v}{2\epsilon_0} \left[ L - \frac{1}{2} \left( \sqrt{(x-x_0)^2 + a^2} - \sqrt{(x-x_0-L)^2 + a^2} \right) \right] } \] --- ## **Summary of Answers** **(a) Disk:** \[ \boxed{ E_x(x,0,0) = \frac{\rho_s}{2\epsilon_0} \left[1 - \frac{x-x_0}{\sqrt{(x-x_0)^2 + a^2}}\right] } \] **(b) Cylinder:** \[ \boxed{ E_x(x,0,0) = \frac{\rho_v}{2\epsilon_0} \left[ L - \frac{1}{2} \left( \sqrt{(x-x_0)^2 + a^2} - \sqrt{(x-x_0-L)^2 + a^2} \right) \right] } \]

Let's carefully work through each part step by step. --- ### **Part (a): Electric field at a point on the axis of a uniformly charged disk** **Given:** - Disk radius: \( a \) - Surface charge density: \( \rho_s \) - Disk is in the \( yz \)-plane at \( x = x_0 \) - Observation point: \( (x, 0, 0) \) along the \( x \)-axis --- #### **Step 1: Set up the problem** Since the disk is symmetric in the \( yz \)-plane, the electric field contribution in the \( y \) and \( z \) directions cancels out, leaving only the \( x \)-component. We consider a small surface element \( dS \) at position \( (x_0, y', z') \). The differential surface charge: \[ dQ = \rho_s \, dS \] The field at \( (x, 0, 0) \) due to this element is: \[ d\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{dQ\, \vec{R}}{|\vec{R}|^3} \] where \( \vec{R} \) points from the charge element to the observation point: \[ \vec{R} = (x - x_0)\hat{x} - y' \hat{y} - z' \hat{z} \] --- #### **Step 2: Symmetry considerations** - Due to symmetry, the \( y \) and \( z \) components cancel when integrated over the entire disk, leaving only the \( x \)-component: \[ E_x = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho_s \, dS \, (x - x_0)}{|\vec{R}|^3} \] - In cylindrical coordinates: \[ dS = r' dr' d\theta \] with \( y' = r' \cos \theta \), \( z' = r' \sin \theta \), and \( 0 \le r' \le a \), \( 0 \le \theta < 2\pi \). - The magnitude: \[ |\vec{R}| = \sqrt{(x - x_0)^2 + r'^2} \] - The projection of \( d\vec{E} \) along \( x \): \[ dE_x = \frac{1}{4 \pi \epsilon_0} \frac{\rho_s r' dr' d\theta (x - x_0)}{\left[ (x - x_0)^2 + r'^2 \right]^{3/2}} \] --- #### **Step 3: Integrate over \( \theta \) and \( r' \)** Since the integrand doesn't depend on \( \theta \): \[ E_x = \frac{\rho_s (x - x_0)}{4 \pi \epsilon_0} \times 2\pi \int_0^a \frac{r' dr'}{\left[ (x - x_0)^2 + r'^2 \right]^{3/2}} \] Simplify: \[ E_x = \frac{\rho_s (x - x_0)}{2 \epsilon_0} \int_0^a \frac{r' dr'}{\left[ (x - x_0)^2 + r'^2 \right]^{3/2}} \] --- #### **Step 4: Evaluate the integral** Let: \[ u = (x - x_0)^2 + r'^2 \quad \Rightarrow \quad du = 2 r' dr' \] When \( r' = 0 \), \( u = (x - x_0)^2 \) When \( r' = a \), \( u = (x - x_0)^2 + a^2 \) The integral becomes: \[ \int_{u = (x - x_0)^2}^{(x - x_0)^2 + a^2} \frac{du/2}{u^{3/2}} = \frac{1}{2} \int_{(x - x_0)^2}^{(x - x_0)^2 + a^2} u^{-3/2} du \] The integral: \[ \int u^{-3/2} du = -2 u^{-1/2} \] Thus: \[ \int_0^a \frac{r' dr'}{\left[ (x - x_0)^2 + r'^2 \right]^{3/2}} = \frac{1}{2} \left[ -2 u^{-1/2} \right]_{(x - x_0)^2}^{(x - x_0)^2 + a^2} = - u^{-1/2} \Big|_{(x - x_0)^2}^{(x - x_0)^2 + a^2} \] Evaluate: \[ = - \left[ \frac{1}{\sqrt{ (x - x_0)^2 + a^2 }} - \frac{1}{|x - x_0|} \right] \] --- #### **Step 5: Final expression for \( E_x \)** Putting it all together: \[ E_x = \frac{\rho_s (x - x_0)}{2 \epsilon_0} \times \left[ \frac{1}{|x - x_0|} - \frac{1}{\sqrt{(x - x_0)^2 + a^2}} \right] \] Note that \( (x - x_0)/|x - x_0| = \pm 1 \), which simplifies to: \[ \boxed{ \textbf{(a)}\quad E_x(x) = \frac{\rho_s}{2 \epsilon_0} \left[ 1 - \frac{x - x_0}{\sqrt{(x - x_0)^2 + a^2}} \right] } \] This is the electric field at point \( x \) on the axis of a uniformly charged disk. --- ### **Part (b): Electric field along the axis of a solid cylinder with volume charge density \( \rho_v \)** The cylinder has: - Radius: \( a \) - Length: \( L \) - Positioned along the \( x \)-axis from \( x = x_0 \) to \( x = x_0 + L \) - Volume charge density: \( \rho_v \) --- #### **Step 1: Model as stacked disks** The volume charge density can be viewed as a stack of infinitesimal disks of thickness \( dx_1 \): \[ dQ = \rho_v \, dV = \rho_v \times \text{area} \times dx_1 = \rho_v \pi a^2 dx_1 \] The surface charge density of each disk: \[ \rho_s = \frac{dQ}{dS} = \frac{\rho_v \pi a^2 dx_1}{\pi a^2} = \rho_v dx_1 \] --- #### **Step 2: Contribution of each disk** From part (a), the field at \( x \) due to a disk at position \( x_1 \): \[ dE_x = \frac{\rho_v dx_1}{2 \epsilon_0} \left[ 1 - \frac{x - x_1}{\sqrt{(x - x_1)^2 + a^2}} \right] \] The total field is obtained by integrating over the length of the cylinder: \[ E_x = \int_{x_0}^{x_0 + L} dE_x \] \[ E_x = \frac{\rho_v}{2 \epsilon_0} \int_{x_0}^{x_0 + L} \left[ 1 - \frac{x - x_1}{\sqrt{(x - x_1)^2 + a^2}} \right] dx_1 \] --- #### **Step 3: Perform the integral** Split into two integrals: \[ E_x = \frac{\rho_v}{2 \epsilon_0} \left[ \int_{x_0}^{x_0 + L} dx_1 - \int_{x_0}^{x_0 + L} \frac{x - x_1}{\sqrt{(x - x_1)^2 + a^2}} dx_1 \right] \] The first integral: \[ \int_{x_0}^{x_0 + L} dx_1 = L \] The second integral: \[ I = \int_{x_0}^{x_0 + L} \frac{x - x_1}{\sqrt{(x - x_1)^2 + a^2}} dx_1 \] Let: \[ u = x - x_1 \quad \Rightarrow \quad du = - dx_1 \] When \( x_1 = x_0 \), \( u = x - x_0 \) When \( x_1 = x_0 + L \), \( u = x - x_0 - L \) Rewriting the integral: \[ I = \int_{u = x - x_0}^{x - x_0 - L} \frac{u}{\sqrt{u^2 + a^2}} (-du) = - \int_{u = x - x_0}^{x - x_0 - L} \frac{u}{\sqrt{u^2 + a^2}} du \] Switch limits: \[ I = \int_{u = x - x_0 - L}^{x - x_0} \frac{u}{\sqrt{u^2 + a^2}} du \] The integral: \[ \int \frac{u}{\sqrt{u^2 + a^2}} du = \sqrt{u^2 + a^2} \] Evaluate: \[ I = \left[ \sqrt{u^2 + a^2} \right]_{u = x - x_0 - L}^{u = x - x_0} \] \[ I = \sqrt{(x - x_0)^2 + a^2} - \sqrt{(x - x_0 - L)^2 + a^2} \] --- #### **Step 4: Final expression** Putting it all together: \[ E_x = \frac{\rho_v}{2 \epsilon_0} \left[ L - \left( \sqrt{(x - x_0)^2 + a^2} - \sqrt{(x - x_0 - L)^2 + a^2} \right) \right] \] Or: \[ \boxed{ \textbf{(b)} \quad E_x(x) = \frac{\rho_v}{2 \epsilon_0} \left[ L - \sqrt{(x - x_0)^2 + a^2} + \sqrt{(x - x_0 - L)^2 + a^2} \right] } \] --- ### **Final Results:** \[ \boxed{ \textbf{(a)} \quad E_x(x) = \frac{\rho_s}{2 \epsilon_0} \left[ 1 - \frac{x - x_0}{\sqrt{(x - x_0)^2 + a^2}} \right] } \] \[ \boxed{ \textbf{(b)} \quad E_x(x) = \frac{\rho_v}{2 \epsilon_0} \left[ L - \sqrt{(x - x_0)^2 + a^2} + \sqrt{(x - x_0 - L)^2 + a^2} \right] } \] These expressions give the electric field along the axis due to the charged disk and the charged cylinder, respectively.