Question:
A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)
A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)
Show Hint
Use the formula for the electric field at the center of a charged semicircular arc: $E = \frac{2k\lambda}{R}$, where $\lambda = \frac{Q}{\pi R}$.
Updated On: Apr 12, 2026
- 2.14
- 2.44
- 3.25
- 0.7
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
To solve this problem, we first determine the expression for the electric field at the center of a uniformly charged semicircular arc.
Let the total charge on the half ring be $Q$ and its radius be $R = 35 \text{ cm} = 0.35 \text{ m}$. The linear charge density $\lambda$ is defined as the charge per unit length. For a semicircular arc of radius $R$, the length is $\pi R$, so:
$$\lambda = \frac{Q}{\pi R}$$
Consider an infinitesimal element of the ring subtending an angle $d\theta$ at the center. The charge on this element is $dq = \lambda dl = \lambda (R d\theta)$. The electric field $dE$ at the center due to this element is:
$$dE = \frac{1}{4\pi\epsilon_o} \frac{dq}{R^2} = \frac{1}{4\pi\epsilon_o} \frac{\lambda R d\theta}{R^2} = \frac{\lambda}{4\pi\epsilon_o R} d\theta$$
Due to symmetry, the horizontal components of the electric field from opposite sides of the ring cancel out. We only need to integrate the vertical components (along the axis of symmetry):
$$E = \int dE \sin\theta = \int_{0}^{\pi} \frac{\lambda}{4\pi\epsilon_o R} \sin\theta d\theta$$
$$E = \frac{\lambda}{4\pi\epsilon_o R} [-\cos\theta]_{0}^{\pi} = \frac{\lambda}{4\pi\epsilon_o R} (1 - (-1)) = \frac{2\lambda}{4\pi\epsilon_o R} = \frac{\lambda}{2\pi\epsilon_o R}$$
Substituting $\lambda = \frac{Q}{\pi R}$ into the formula:
$$E = \frac{Q}{2\pi^2 \epsilon_o R^2}$$
Given $E = 100 \text{ V/m}$, $R = 0.35 \text{ m}$, $\pi = 3.14$, and $\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$. Rearranging for $Q$:
$$Q = E \cdot 2\pi^2 \epsilon_o R^2$$
$$Q = 100 \times 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2$$
$$Q = 200 \times 9.8596 \times 8.85 \times 0.1225 \times 10^{-12}$$
$$Q \approx 2137.8 \times 10^{-12} \text{ C} = 2.1378 \times 10^{-9} \text{ C}$$
Since $1 \text{ nC} = 10^{-9} \text{ C}$, the value is approximately $2.14 \text{ nC}$.
Let the total charge on the half ring be $Q$ and its radius be $R = 35 \text{ cm} = 0.35 \text{ m}$. The linear charge density $\lambda$ is defined as the charge per unit length. For a semicircular arc of radius $R$, the length is $\pi R$, so:
$$\lambda = \frac{Q}{\pi R}$$
Consider an infinitesimal element of the ring subtending an angle $d\theta$ at the center. The charge on this element is $dq = \lambda dl = \lambda (R d\theta)$. The electric field $dE$ at the center due to this element is:
$$dE = \frac{1}{4\pi\epsilon_o} \frac{dq}{R^2} = \frac{1}{4\pi\epsilon_o} \frac{\lambda R d\theta}{R^2} = \frac{\lambda}{4\pi\epsilon_o R} d\theta$$
Due to symmetry, the horizontal components of the electric field from opposite sides of the ring cancel out. We only need to integrate the vertical components (along the axis of symmetry):
$$E = \int dE \sin\theta = \int_{0}^{\pi} \frac{\lambda}{4\pi\epsilon_o R} \sin\theta d\theta$$
$$E = \frac{\lambda}{4\pi\epsilon_o R} [-\cos\theta]_{0}^{\pi} = \frac{\lambda}{4\pi\epsilon_o R} (1 - (-1)) = \frac{2\lambda}{4\pi\epsilon_o R} = \frac{\lambda}{2\pi\epsilon_o R}$$
Substituting $\lambda = \frac{Q}{\pi R}$ into the formula:
$$E = \frac{Q}{2\pi^2 \epsilon_o R^2}$$
Given $E = 100 \text{ V/m}$, $R = 0.35 \text{ m}$, $\pi = 3.14$, and $\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$. Rearranging for $Q$:
$$Q = E \cdot 2\pi^2 \epsilon_o R^2$$
$$Q = 100 \times 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2$$
$$Q = 200 \times 9.8596 \times 8.85 \times 0.1225 \times 10^{-12}$$
$$Q \approx 2137.8 \times 10^{-12} \text{ C} = 2.1378 \times 10^{-9} \text{ C}$$
Since $1 \text{ nC} = 10^{-9} \text{ C}$, the value is approximately $2.14 \text{ nC}$.
Was this answer helpful?
1
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
View More Questions
Top JEE Main Electric Field Questions
- The distance between charges \(+q\) and \(-q\) is \(2l\) and between \(+2q\) and \(-2q\) is \(4l\). The electrostatic potential at point \(P\) at a distance \(r\) from center \(O\) is \(-\alpha \left[\frac{q}{r^2}\right] \times 10^9 \, \text{V}\), where the value of \(\alpha\) is ______.
(Use \(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}\))
- \(\sigma\) is the uniform surface charge density of a thin spherical shell of radius \(R\). The electric field at any point on the surface of the spherical shell is:
- An electron of mass \( m \) with an initial velocity \( \vec{v} = v_0 \hat{i} (v_0>0) \) enters an electric field \( \vec{E} = -E_0 \hat{k} \). If the initial de Broglie wavelength is \( \lambda_0 \), the value after time \( t \) would be:
- An electron is released in the field generated by a non-conductivity sheet of uniform surface charge density $ \sigma $. The rate of change of de-Broglie wavelength associated with the electron varies inversely as the $ n $th power of the distance travelled. Find the value of $ n $.
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC?
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions