A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to 'O' is 'E' is magnitude. What would be the magnitude of electric field at 'O' due to arc ABC?

### Related Formula The electric field due to a circular arc subtending an angle phi$\phi$ at the center is given by: E_textarc = frac2klambdaR sinleft(fracphi2right)$$E_{\text{arc}} = \frac{2k\lambda}{R} \sin\left(\frac{\phi}{2}\right)$$ ### Core Logic Arc AB subtends 90^circ$90^\circ$ (one quadrant) at the center. The electric field due to it is given as E$E$. Arc ABC consists of two independent quadrants: arc AB and arc BC. Each quadrant independently creates an electric field of magnitude E$E$ pointing along the bisector of that specific quadrant. ### Step 1: Vector Addition The electric field vecE_AB$\vec{E}_{AB}$ is directed at 45^circ$45^\circ$ away from both axes into the third quadrant. The electric field vecE_BC$\vec{E}_{BC}$ is directed at 45^circ$45^\circ$ towards the matching opposite quadrant. Since vecE_AB$\vec{E}_{AB}$ and vecE_BC$\vec{E}_{BC}$ are perpendicular to each other, their resultant magnitude is: E_textnet = sqrtE^2 + E^2 = sqrt2E$$E_{\text{net}} = \sqrt{E^2 + E^2} = \sqrt{2}E$$

### Pattern Recognition Symmetric components of a ring create orthogonal vector fields. Each 90^circ$90^\circ$ arc produces a field of magnitude E$E$ directed along its angular bisector. Two adjacent quadrants have bisectors separated by 90^circ$90^\circ$, hence use orthogonal vector addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics