Q.38. A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the

Q: What is the correct answer for JEE Main Physics Q38?

The correct answer is A. Given: Radius of half ring $R = 35$ cm $= 0.35$ m Electric field at the center $E = 100$ V/m Total charge on the half ring = Q Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ C²/Nm² $\pi = 3.14$ For a uniformly charged half ring, the electric field at its center is given by the formula: $E = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R}$, where $\lambda$ is the linear charge density.

Q: Q.38. 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 ____…

The correct answer is A. Given: Radius of half ring $R = 35$ cm $= 0.35$ m Electric field at the center $E = 100$ V/m Total charge on the half ring = Q Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ C²/Nm² $\pi = 3.14$ For a uniformly charged half ring, the electric field at its center is given by the formula: $E = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R}$, where $\lambda$ is the linear charge density.

Q: What are the options for this JEE Main Physics question?

A. 14 · B. 44 · C. 25 · D. 7

Question

Q.38. 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. ( C²/Nm² and )

FirstInTest · Accepted Answer

Given:
Radius of half ring $R = 35$ cm $= 0.35$ m
Electric field at the center $E = 100$ V/m
Total charge on the half ring = Q
Permittivity of free space $\epsilon_0 = 8.85 	imes 10^{-12}$ C²/Nm²
$\pi = 3.14$

For a uniformly charged half ring, the electric field at its center is given by the formula:
$E = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R}$, where $\lambda$ is the linear charge density.

The linear charge density $\lambda = \frac{	ext{Total Charge}}{	ext{Length of half ring}}$.
The length of a half ring is $\pi R$.
So, $\lambda = \frac{Q}{\pi R}$.

Substitute $\lambda$ into the electric field formula:
$E = \frac{1}{4\pi\epsilon_0} \frac{2(Q/\pi R)}{R} = \frac{1}{4\pi\epsilon_0} \frac{2Q}{\pi R^2}$

We know that $k = \frac{1}{4\pi\epsilon_0} = 9 	imes 10^9$ N m²/C².
So, $E = k \frac{2Q}{\pi R^2}$.

Rearrange to solve for Q:
$Q = \frac{E \pi R^2}{2k}$

Substitute the given values:
$E = 100$ V/m
$R = 0.35$ m
$k = 9 	imes 10^9$ N m²/C²
$\pi = 3.14$

$Q = \frac{100 	imes 3.14 	imes (0.35)^2}{2 	imes 9 	imes 10^9}$
$Q = \frac{100 	imes 3.14 	imes 0.1225}{18 	imes 10^9}$
$Q = \frac{314 	imes 0.1225}{18 	imes 10^9} = \frac{38.465}{18 	imes 10^9} \approx 2.1369 	imes 10^{-9}$ C

To express Q in nC (nanocoulombs), multiply by $10^9$:
$Q \approx 2.1369$ nC.
Rounding to two decimal places, $Q \approx 2.14$ nC.

Answer

14

Answer

44

Answer

25

Answer

7

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