### Related Formula 1. Electric field intensity on the axis of a ring of radius R$R$ at axial distance z$z$: E = frack Q zleft(z^2 + R^2right)^3/2$E = \frac{k Q z}{\left(z^2 + R^2\right)^{3/2}}$ 2. Condition for maximum electric field on the axis of a ring: z = fracRsqrt2$z = \frac{R}{\sqrt{2}}$ ### Core Logic To find where the axial field is maximized, take the first derivative of E$E$ with respect to z$z$ and set it to zero: fracdEdz = 0 implies z^2 + R^2 - 3z^2 = 0 implies z = fracRsqrt2$\frac{dE}{dz} = 0 \implies z^2 + R^2 - 3z^2 = 0 \implies z = \frac{R}{\sqrt{2}}$ We are given: - Radius of the loop R = asqrt2$R = a\sqrt{2}$ ### Step 1: Calculate the peak coordinate Substitute the radius R = asqrt2$R = a\sqrt{2}$ into the maximized coordinate condition: z = fracasqrt2sqrt2 = a$z = \frac{a\sqrt{2}}{\sqrt{2}} = a$ Thus, the electric field reaches its peak intensity at z = a$z = a$. ### Pattern Recognition Sees: Maximum axial field location of circular charged ring. Trap: Simply picking radius R$R$ directly as the distance value, or omitting the 1/sqrt2$1/\sqrt{2}$ scaling factor. Shortcut: The peak of the axial field distribution of any ring always lies at z = fracRsqrt2$z = \frac{R}{\sqrt{2}}$. With R = asqrt2$R = a\sqrt{2}$, z$z$ resolves simply to fracasqrt2sqrt2 = a$\frac{a\sqrt{2}}{\sqrt{2}} = a$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electric Charges and Fields

### Core Logic Applying the work-energy balance for the system as the particle moves from its initial position to the x-axis: mathrmW_textall = Delta mathrmk$\mathrm{W}_{\text{all}} = \Delta \mathrm{k}$ mathrmW_mathrme = mathrmk_mathrmf - mathrmk_mathrmi$\mathrm{W}_{\mathrm{e}} = \mathrm{k}_{\mathrm{f}} - \mathrm{k}_{\mathrm{i}}$

The displacement parallel to the uniform electric field lines equals fracell2$\frac{\ell}{2}$. Therefore, computing the work done by the electrostatic field: mathrmqE fracell2 = frac12 mathrmm v^2 - 0$\mathrm{qE} \frac{\ell}{2} = \frac{1}{2} \mathrm{m v}^2 - 0$ ### Step 1: Final Expression mathrmv = sqrtfracmathrmqEellmathrmm$\mathrm{v} = \sqrt{\frac{\mathrm{qE}\ell}{\mathrm{m}}}$ This matches option (3). ### Pattern Recognition Keep your focus on displacement along the field lines. Work relies entirely on the parallel displacement component (d_x = ell cos 60^circ$d_x = \ell \cos 60^{\circ}$), completely ignoring any vertical movement components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electric Charges and Fields

### Related Formula Gauss's Law states: Phi = oint vecE cdot dvecA = fracq_textenclosedvarepsilon_0$\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$ ### Core Logic We have a sphere of radius R = 4a$R = 4a$ centered at (0,0)$(0,0)$. * Charge 5Q$5Q$ is located at (3a, 0)$(3a, 0)$. Since the distance from origin is 3a lt 4a$3a \lt 4a$, this charge lies **inside** the sphere. * Charge -2Q$-2Q$ is located at (-5a, 0)$(-5a, 0)$. Since the distance from origin is 5a gt 4a$5a \gt 4a$, this charge lies **outside** the sphere.

### Step 1: Calculate Enclosed Charge The net enclosed charge q_textenclosed$q_{\text{enclosed}}$ within the spherical boundary is: q_textenclosed = 5Q$q_{\text{enclosed}} = 5Q$ ### Step 2: Apply Gauss Law Using Gauss's Law, the total electric flux is: Phi = fracq_textenclosedvarepsilon_0 = frac5Qvarepsilon_0$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{5Q}{\varepsilon_0}$ Therefore, the electric flux is frac5Qvarepsilon_0$\frac{5Q}{\varepsilon_0}$. ### Pattern Recognition Flux depends purely on charges situated *inside* the closed surface. Charges outside the Gaussian surface contribute absolutely zero net flux because every field line entering must also exit. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electric Charges and Fields

### Related Formula The uniform electric field (E$E$) produced by an infinite plane sheet with positive charge density +sigma$+\sigma$ is: E = fracsigma2varepsilon_0$E = \frac{\sigma}{2\varepsilon_0}$ The constant electrostatic force acting on an electron (charge -e$-e$, mass m$m$) directed towards the sheet is: F = e E = fracsigma e2varepsilon_0 implies textAcceleration a = -fracsigma e2varepsilon_0 m$F = e E = \frac{\sigma e}{2\varepsilon_0} \implies \text{Acceleration } a = -\frac{\sigma e}{2\varepsilon_0 m}$ ### Core Logic For the boundary limits matching a successful collision with the plate under maximum field conditions, the electron must be moving directly away from sheet $ ### Core Logic For the boundary limits matching a successful collision with the plate under maximum field conditions, the electron must be moving directly away from sheet $S initially ($ initially ($u = +1 \mathrm{~m/s}). The accelerating field acts as a decelerating force, turning it around to hit the sheet exactly at $). The accelerating field acts as a decelerating force, turning it around to hit the sheet exactly at $t = 1 \mathrm{~s}. Therefore, relative to the initial position vector pointing outwards: * Initial speed $. Therefore, relative to the initial position vector pointing outwards: * Initial speed $u = 1 \mathrm{~m/s} * Net displacement at strike $ * Net displacement at strike $S = -1 \mathrm{~m} (since it returns to the plate at origin) * Total time elapsed $ (since it returns to the plate at origin) * Total time elapsed $t = 1 \mathrm{~s} ### Step 1: Apply Second Equation of Motion Using $ ### Step 1: Apply Second Equation of Motion Using $S = u t + \frac{1}{2} a t^2: $: $-1 = 1 times 1 + frac12 a (1)^2$-1 = 1 \times 1 + \frac{1}{2} a (1)^2$ $ $-1 = 1 + frac12 a implies frac12 a = -2 implies a = -4 mathrm~m/s^2$-1 = 1 + \frac{1}{2} a \implies \frac{1}{2} a = -2 \implies a = -4 \mathrm{~m/s^2}$ ### Step 2: Solve for Surface Charge Density Equating this deceleration value to the electrostatic tracking acceleration: $ ### Step 2: Solve for Surface Charge Density Equating this deceleration value to the electrostatic tracking acceleration: $-4 = -fracsigma e2varepsilon_0 m$-4 = -\frac{\sigma e}{2\varepsilon_0 m}$ $ $sigma = 8 fracvarepsilon_0 me = 8 left[ fracm varepsilon_0e right]$\sigma = 8 \frac{\varepsilon_0 m}{e} = 8 \left[ \frac{m \varepsilon_0}{e} \right]$ ### Step 3: Extract alpha Comparing this with the algebraic pattern $ ### Step 3: Extract alpha Comparing this with the algebraic pattern $\alpha \left[\frac{m \varepsilon_0}{e}\right]: $: $\alpha = 8 Therefore, the value of $ Therefore, the value of $\alpha is $ is $8. ### Pattern Recognition Be careful with coordinate signs. If the electron were moving towards the plate initially ($. ### Pattern Recognition Be careful with coordinate signs. If the electron were moving towards the plate initially ($u = -1), any field value would pull it in even faster, meaning the time would be less than $), any field value would pull it in even faster, meaning the time would be less than $1 \mathrm{~s}$. The 'maximum field' boundary constraint implies the electron is thrown away and turned around at its apex peak, mapping perfectly to a return displacement. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electric Charges and Fields Class 11 Physics: Motion in a Straight Line