Three charges +2q$+2q$, +3q$+3q$ and -4q$-4q$ are situated at (0,-3a)$(0,-3a)$, (2a, 0)$(2a, 0)$ and (-2a,0)$(-2a,0)$ respectively in the xy plane. The resultant dipole moment about origin is

### 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 The electric flux Phi$\Phi$ through an area vector vecA$\vec{A}$ in a uniform electric field vecE$\vec{E}$ is: Phi = vecE cdot vecA$$\Phi = \vec{E} \cdot \vec{A}$$ ### Core Logic Given: * vecE = 6hati + 5hatj + 3hatk$\vec{E} = 6\hat{i} + 5\hat{j} + 3\hat{k}$ * vecA = 30hati$\vec{A} = 30\hat{i}$ (since the surface lies in the YZ$YZ$-plane, its normal area vector points along the X$X$-axis, which is hati$\hat{i}$) ### Step 1: Compute Dot Product Using the vector dot product: Phi = (6hati + 5hatj + 3hatk) cdot (30hati)$$\Phi = (6\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (30\hat{i})$$ Phi = (6 times 30) + 0 + 0 = 180text N m^2/textC$$\Phi = (6 \times 30) + 0 + 0 = 180\text{ N m}^2/\text{C}$$ Thus, the flux is 180$180$. ### Pattern Recognition Flux through a plane parallel to YZ$YZ$-plane depends only on the x$x$-component of the electric field (E_x$E_x$). Therefore, Phi = E_x times A = 6 times 30 = 180$\Phi = E_x \times A = 6 \times 30 = 180$ immediately. ### 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