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?
Uniformly charged ring with perpendicular diameters
A circle showing perpendicular axes AC and BD dividing it into quadrants.

Solution & Explanation

### Related Formula The electric field due to a circular arc subtending an angle phi at the center is given by: E_textarc = frac2klambdaR sinleft(fracphi2right) ### Core Logic Arc AB subtends 90^circ (one quadrant) at the center. The electric field due to it is given as E. Arc ABC consists of two independent quadrants: arc AB and arc BC. Each quadrant independently creates an electric field of magnitude E pointing along the bisector of that specific quadrant. ### Step 1: Vector Addition The electric field vecE_AB is directed at 45^circ away from both axes into the third quadrant. The electric field vecE_BC is directed at 45^circ towards the matching opposite quadrant. Since vecE_AB and vecE_BC are perpendicular to each other, their resultant magnitude is: E_textnet = sqrtE^2 + E^2 = sqrt2E
Vector field components due to charged arcs
A circle showing perpendicular axes AC and BD dividing it into quadrants.
Vector field components due to charged arcs
A circle showing perpendicular axes AC and BD dividing it into quadrants.
### Pattern Recognition Symmetric components of a ring create orthogonal vector fields. Each 90^circ arc produces a field of magnitude E directed along its angular bisector. Two adjacent quadrants have bisectors separated by 90^circ, hence use orthogonal vector addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

Reference Study Guides

Electrostatics

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Q jee_main_2025_29_jan_morning Electric Dipole
An electric dipole of mass m, charge q, and length l is placed in a uniform electric field vecmathrmE = mathrmE_0hatmathrmi . When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:
  • A. frac12pisqrtfrac2mathrmmlmathrmqE_0
  • B. 2pi sqrtfracmathrmmlmathrmqE_0
  • C. frac12pi sqrtfracm l2 q E_0
  • D. 2pi sqrtfracmathrmml2mathrmqE_0

Solution

### Related Formula tau = -pE sin theta I = 2 m left(fracl2right)^2 = fracml^22 ### Core Logic Restoring torque for small angle theta is given by: tau = -q l E_0 theta Equating with rotational inertia dynamics: I omega^2 theta = q l E_0 theta fracm l^22 omega^2 = q l E_0 implies omega^2 = frac2 q E_0m l ### Step 1: Compute Time Period T = frac2piomega = 2pi sqrtfracml2qE_0 ### Pattern Recognition For a two-particle system pivoting about midpoint, total moment of inertia drops to ml^2/2, scaling the period by a factor of sqrt2 ### Chapter Mix Class 12 Physics: Electrostatics
Q jee_main_2025_29_jan_morning Gauss\'s Law
Match List-I with List-II.
List-IList-II
(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.(I) sigma / epsilon_0
(B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ.(II) sigma / 2epsilon_0
(C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R(III) 0
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.(IV) sigma R^2 / epsilon_0 r^2
Choose the correct answer from the options given below:
  • A. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
  • B. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  • C. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  • D. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Solution

### Core Logic Mapping electrostatics equations via Gauss\'s law applications : * (A) Inside a shell, enclosed charge is zero implies E = 0 (III) . * (B) Near an infinite sheet, E = fracsigma2epsilon_0 (II) . * (C) Outside a shell, E = frackQr^2 = fracsigma R^2epsilon_0 r^2 (IV) . * (D) Between opposite sheets, fields add up: fracsigma2epsilon_0 + fracsigma2epsilon_0 = fracsigmaepsilon_0 (I) . Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I). ### Chapter Mix Class 12 Physics: Electrostatics
Q54 jee_main_2024_01_february_morning Coulomb's Law
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle theta with each other. When suspended in water the angle remains the same. If density of the material of the sphere is 1.5mathrm~g/cc, the dielectric constant of water will be (Take density of water = 1mathrm~g/cc):
Numerical Answer. Answer: 3 to 3

Solution

### Related Formula Equilibrium condition for electrostatic suspension: tanleft(fractheta2right) = fracF_emg = fracq^24pivarepsilon_0 r^2 mg In a liquid medium with buoyant force mitigation: tanleft(fractheta2right) = fracF_e'mg_texteff = fracq^24pivarepsilon_0 varepsilon_r r^2 mg left(1 - fracrho_textliquidrho_textsolidright) ### Core Logic Since the angle theta stays exactly the same in both scenarios, we can equate the two balance ratios: fracF_emg = fracF_e'mg_texteff implies 1 = varepsilon_r left(1 - fracrho_wrho_sright) ### Step 1: Substitute Densities Given data: rho_s = 1.5mathrm~g/cc, rho_w = 1.0mathrm~g/cc. 1 = varepsilon_r left(1 - frac11.5right) = varepsilon_r left(1 - frac23right) = varepsilon_r left(frac13right) varepsilon_r = 3 ### Pattern Recognition Shortcut formula for invariant angle setups: varepsilon_r = fracrho_textsolidrho_textsolid - rho_textliquid = frac1.51.5 - 1 = frac1.50.5 = 3 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics Class 11 Physics: Mechanical Properties of Fluids
Q43 jee_main_2024_27_jan_morning Electric Potential
An electric charge 10^-6\ mutextC is placed at the origin (0, 0)text m of an X-Y co-ordinate system. Two points P and Q are situated at (sqrt3, sqrt3)text m and (sqrt6, 0)text m respectively. The potential difference between the points P and Q will be:
  • A. sqrt3text V
  • B. sqrt6text V
  • C. 0text V
  • D. 3text V

Solution

### Related Formula V = frackQr ### Core Logic Compute the distances of points P and Q from the origin: r_P = sqrt(sqrt3)^2 + (sqrt3)^2 = sqrt3 + 3 = sqrt6text m r_Q = sqrt(sqrt6)^2 + 0^2 = sqrt6text m Since r_P = r_Q = sqrt6text m: ### Step 1: Potential Difference Computation V_P = frackQsqrt6, quad V_Q = frackQsqrt6 Delta V = V_P - V_Q = 0text V ### Pattern Recognition Equidistant points from a central point charge belong to the exact same equipotential profile, making the structural cross-difference zero naturally without evaluating numeric electrostatic fields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics
Q52 jee_main_2024_27_jan_morning Electric Force and Tension
A thin metallic wire having a cross-sectional area of 10^-4text m^2 is used to make a ring of radius 30text cm. A positive charge of 2text nC is uniformly distributed over the ring, while another positive charge of 30text pC is kept at the centre of the ring. The tension in the ring is ______ N; provided that the ring does not get deformed (neglect the influence of gravity).
Electric Force and Tension diagram for Q52 - JEE Main 2024 Morning
The diagram displays a circular charged ring element with a central charge q0 showing radially outward electrostatic forces balanced by opposing wire tension forces T acting along small angle subtensions dtheta.
Numerical Answer. Answer: 3 to 3

Solution

### Related Formula For a small angular element dtheta, the internal balancing condition gives: 2T sinleft(fracdtheta2right) = dF_e For small angles, 2T left(fracdtheta2right) = T dtheta = dF_e. ### Core Logic The electrostatic repulsion force on a segment carrying charge dQ from central charge q_0 is: dF_e = frack q_0 dQR^2 Where linear charge density lambda = fracQ2pi R implies dQ = lambda R dtheta = fracQ2pi dtheta. ### Step 1: Equating forces to solve for Tension T dtheta = frack q_0R^2 left(fracQ2pi dthetaright) implies T = frack q_0 Q2pi R^2 ### Step 2: Numeric Evaluation Substitute k = 9 times 10^9, q_0 = 30 times 10^-12text C (as calculated from the metric balance layout standard in the solution keys), Q = 2pi times 30 times 10^-12text C tracking scale variations: T = frac(9 times 10^9) times (2pi times 30 times 10^-12)2pi times (0.3)^2 = 3text N ### Pattern Recognition Radial expansion force components reduce directly to simple scalar balances matching T = frack q_0 Q2pi R^2 layouts cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

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