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Physics Question 38 – JEE-MAIN 2026

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A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value of Q is _______ nC. (ϵ0=8.85×10−12 C2/Nm2 and π=3.14)

Recall the formula for the electric field at the center of a uniformly charged half-ring.

Step 1: Identify the relevant formula✦ Active

The electric field at the center of a uniformly charged half-ring of radius R and total charge Q is given by:

E=Q2π2ϵ0R2
Step 2: Rearrange the formula and substitute values○ Expand

Rearrange the formula to solve for Q: Q=E⋅2π2ϵ0R2. Convert the radius to meters: R=35 cm=0.35 m. Substitute the given values: E=100 V/m, ϵ0=8.85×10−12 C2/Nm2, and π=3.14.

Q=100⋅2⋅(3.14)2⋅(8.85×10−12)⋅(0.35)2
Step 3: Calculate the charge and convert units○ Expand

Perform the calculation:

Q=2137.566595×10−12 C

Convert the result from Coulombs to nanoCoulombs (nC), where 1 nC=10−9 C:

Q≈2.1375×10−9 C≈2.14 nC

Thus, the value of Q is approximately 2.14 nC.

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