Physics Q38 – JEE-MAIN 2025

Home Physics Electric Charges and FieldsQuestion #115

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

Clue 1 of 3: Electric Field of an Arc

Recall the formula for the electric field at the center of a uniformly charged circular arc.

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Ninja Strategy•Symmetry and Magnitude Reasoning

Recognize that the field from a semi-circle will be non-zero and likely larger than that from a quadrant, allowing for the elimination of options like 'Zero' and 'E/2'.

🧠 Full Solution Path

Step 1: Identify Arc Angles and Formula✦ Active

Arc AB is a quadrant, subtending an angle of $θ_{A B} = 90^{\circ} = \frac{π}{2}$ radians at the center O. Arc ABC is a semi-circle, subtending an angle of $θ_{A B C} = 180^{\circ} = π$ radians at the center O. The magnitude of the electric field at the center of a uniformly charged arc is given by $E_{a r c} = \frac{2 k λ}{R} \sin (\frac{θ}{2})$ , where $k$ is Coulomb's constant, $λ$ is the linear charge density, and $R$ is the radius.

Step 2: Calculate Field for Arc AB○ Expand

For arc AB, the electric field magnitude at O is given as $E$ . Using the formula:

E = E_{A B} = \frac{2 k λ}{R} \sin (\frac{π / 2}{2}) = \frac{2 k λ}{R} \sin (\frac{π}{4})

E = \frac{2 k λ}{R} (\frac{1}{\sqrt{2}}) = \frac{\sqrt{2} k λ}{R}

From this, we can express $\frac{k λ}{R}$ in terms of $E$ : $\frac{k λ}{R} = \frac{E}{\sqrt{2}}$ .

💡 Teacher's Secret Hint

Ensure correct angle conversion to radians for trigonometric functions.

Step 3: Calculate Field for Arc ABC○ Expand

For arc ABC, the angle is $θ_{A B C} = π$ . Using the formula:

E_{A B C} = \frac{2 k λ}{R} \sin (\frac{π}{2}) = \frac{2 k λ}{R} (1) = \frac{2 k λ}{R}

Now, substitute the expression for $\frac{k λ}{R}$ from Step 2:

E_{A B C} = 2 (\frac{E}{\sqrt{2}}) = \frac{2}{\sqrt{2}} E = \sqrt{2} E

💡 Teacher's Secret Hint

Remember that $\sin (π / 2) = 1$ .

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Physics Question 38 – JEE-MAIN 2025

Clue 1 of 3: Electric Field of an Arc

🧠 Full Solution Path