Question

A point charge q is placed at a distance R from the center O of a uniformly charged hemispherical shell of surface charge density $\sigma$ and radius R on its axis of symmetry as shown in the figure. Then the net electric force experienced by the point charge q due to the uniformly charged hemispherical shell is

• A. $\frac{q\sigma}{\sqrt{2}\epsilon_0}(\sqrt{2}-1)$
• B. $\frac{q\sigma}{4\sqrt{2}\epsilon_0}(\sqrt{2}-1)$
• C. $\frac{q\sigma}{2\sqrt{2}\epsilon_0}(\sqrt{2}-1)$
• D. $\frac{q\sigma}{3\sqrt{2}\epsilon_0}(\sqrt{2}-1)$

Answer 1

Answer: (C)

(3)

Answer 2

To find the net electric force experienced by the point charge $q$ due to the uniformly charged hemispherical shell, we first need to calculate the electric field produced by the hemispherical shell at the location of the point charge.

Let the center of the hemispherical shell be at the origin O (0,0,0). Let the axis of symmetry be the x-axis. The hemispherical shell is on the positive x-axis side, meaning it extends from $x=0$ to $x=R$. The point charge $q$ is located at $(-R, 0, 0)$.

Consider a differential ring element on the hemispherical shell. Let this ring be at an angle $\theta$ with respect to the positive x-axis. The radius of this ring is $r = R\sin\theta$. The width of the ring is $R d\theta$. The area of this differential ring element is $dA = (2\pi R\sin\theta)(R d\theta) = 2\pi R^2 \sin\theta d\theta$. The charge on this ring element is $dQ = \sigma dA = 2\pi R^2 \sigma \sin\theta d\theta$.

A point on this ring can be represented by coordinates $(R\cos\theta, R\sin\theta\cos\phi, R\sin\theta\sin\phi)$. The position of the point charge is $(-R, 0, 0)$. The distance $r'$ from any point on the ring to the point charge $q$ is: $r'^2 = (R\cos\theta - (-R))^2 + (R\sin\theta\cos\phi - 0)^2 + (R\sin\theta\sin\phi - 0)^2$ $r'^2 = (R(1+\cos\theta))^2 + R^2\sin^2\theta(\cos^2\phi+\sin^2\phi)$ $r'^2 = R^2(1+2\cos\theta+\cos^2\theta) + R^2\sin^2\theta$ $r'^2 = R^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta)$ $r'^2 = R^2(2+2\cos\theta) = 2R^2(1+\cos\theta)$. Using the identity $1+\cos\theta = 2\cos^2(\theta/2)$, we get: $r'^2 = 2R^2(2\cos^2(\theta/2)) = 4R^2\cos^2(\theta/2)$. So, $r' = 2R\cos(\theta/2)$. (Since for the hemisphere, $\theta$ ranges from $0$ to $\pi/2$, $\theta/2$ ranges from $0$ to $\pi/4$, and $\cos(\theta/2)$ is positive).

The electric field $d\vec{E}$ due to this ring element at the point charge $q$ will have components perpendicular to the x-axis that cancel out due to symmetry when integrated over the entire ring. Only the x-component of the electric field will contribute to the net field.

The x-component of the electric field $dE_x$ due to a point charge $dQ$ on the ring is given by $dE_x = \frac{1}{4\pi\epsilon_0} \frac{dQ}{r'^2} \cos\alpha$, where $\alpha$ is the angle between the vector from $dQ$ to $q$ and the positive x-axis. The vector from a point on the ring $(R\cos\theta, R\sin\theta, 0)$ to $q(-R, 0, 0)$ is $(-R-R\cos\theta, -R\sin\theta, 0)$. The x-component of this vector is $-R(1+\cos\theta)$. So, $\cos\alpha = \frac{-R(1+\cos\theta)}{r'} = \frac{-R(2\cos^2(\theta/2))}{2R\cos(\theta/2)} = -\cos(\theta/2)$. The negative sign indicates that the electric field is directed towards the negative x-axis (towards the hemisphere).

Now, substitute $dQ$, $r'^2$, and $\cos\alpha$ into the expression for $dE_x$: $dE_x = \frac{1}{4\pi\epsilon_0} \frac{2\pi R^2 \sigma \sin\theta d\theta}{4R^2\cos^2(\theta/2)} (-\cos(\theta/2))$ $dE_x = -\frac{1}{4\pi\epsilon_0} \frac{2\pi R^2 \sigma \sin\theta d\theta}{4R^2\cos(\theta/2)}$ $dE_x = -\frac{\sigma}{8\epsilon_0} \frac{\sin\theta}{\cos(\theta/2)} d\theta$. Using the identity $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$: $dE_x = -\frac{\sigma}{8\epsilon_0} \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos(\theta/2)} d\theta$ $dE_x = -\frac{\sigma}{4\epsilon_0} \sin(\theta/2) d\theta$.

To find the total electric field $E_x$, we integrate $dE_x$ over the entire hemisphere. The angle $\theta$ ranges from $0$ (the tip of the hemisphere, $x=R$) to $\pi/2$ (the base of the hemisphere, $x=0$). $E_x = \int_0^{\pi/2} -\frac{\sigma}{4\epsilon_0} \sin(\theta/2) d\theta$. Let $u = \theta/2$, so $du = \frac{1}{2}d\theta \Rightarrow d\theta = 2du$. When $\theta=0$, $u=0$. When $\theta=\pi/2$, $u=\pi/4$. $E_x = \int_0^{\pi/4} -\frac{\sigma}{4\epsilon_0} \sin(u) (2du)$ $E_x = -\frac{\sigma}{2\epsilon_0} \int_0^{\pi/4} \sin(u) du$ $E_x = -\frac{\sigma}{2\epsilon_0} [-\cos(u)]_0^{\pi/4}$ $E_x = \frac{\sigma}{2\epsilon_0} [\cos(u)]_0^{\pi/4}$ $E_x = \frac{\sigma}{2\epsilon_0} (\cos(\pi/4) - \cos(0))$ $E_x = \frac{\sigma}{2\epsilon_0} \left(\frac{1}{\sqrt{2}} - 1\right)$ $E_x = -\frac{\sigma}{2\epsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right) = -\frac{\sigma}{2\epsilon_0} \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$ $E_x = -\frac{\sigma}{2\sqrt{2}\epsilon_0} (\sqrt{2}-1)$.

The net electric force experienced by the point charge $q$ is $F = qE_x$. $F = q \left(-\frac{\sigma}{2\sqrt{2}\epsilon_0} (\sqrt{2}-1)\right)$. The magnitude of the force is $F = \frac{q\sigma}{2\sqrt{2}\epsilon_0} (\sqrt{2}-1)$.

Comparing this result with the given options: (1) $\frac{q\sigma}{\sqrt{2}\epsilon_0}(\sqrt{2}-1)$ (2) $\frac{q\sigma}{4\sqrt{2}\epsilon_0}(\sqrt{2}-1)$ (3) $\frac{q\sigma}{2\sqrt{2}\epsilon_0}(\sqrt{2}-1)$ (4) $\frac{q\sigma}{3\sqrt{2}\epsilon_0}(\sqrt{2}-1)$

The calculated magnitude matches option (3).

The final answer is $\boxed{\text{(3)}}$

Solution:

1. Define Coordinate System and Elements: Place the center of the hemispherical shell at the origin (0,0,0). The axis of symmetry is the x-axis, with the hemisphere extending for $x \ge 0$. The point charge $q$ is at $(-R, 0, 0)$. Consider a differential ring element on the hemisphere at angle $\theta$ from the positive x-axis.

   • Area of ring $dA = 2\pi R^2 \sin\theta d\theta$.
   • Charge on ring $dQ = \sigma dA = 2\pi R^2 \sigma \sin\theta d\theta$.

2. Distance and Angle: The distance $r'$ from any point on the ring to the point charge $q$ is $r' = 2R\cos(\theta/2)$. The angle $\alpha$ between the vector from $dQ$ to $q$ and the positive x-axis is such that $\cos\alpha = -\cos(\theta/2)$.

3. Electric Field Contribution: By symmetry, only the x-component of the electric field contributes.

   $dE_x = \frac{1}{4\pi\epsilon_0} \frac{dQ}{r'^2} \cos\alpha$

   Substitute $dQ$, $r'$, and $\cos\alpha$:

   $dE_x = \frac{1}{4\pi\epsilon_0} \frac{2\pi R^2 \sigma \sin\theta d\theta}{(2R\cos(\theta/2))^2} (-\cos(\theta/2))$

   $dE_x = -\frac{\sigma}{4\epsilon_0} \sin(\theta/2) d\theta$.

4. Integration: Integrate $dE_x$ from $\theta=0$ to $\theta=\pi/2$ (for the hemisphere):

   $E_x = \int_0^{\pi/2} -\frac{\sigma}{4\epsilon_0} \sin(\theta/2) d\theta$

   Let $u = \theta/2$, $du = (1/2)d\theta$. Limits: $0 \to \pi/4$.

   $E_x = -\frac{\sigma}{2\epsilon_0} \int_0^{\pi/4} \sin(u) du = -\frac{\sigma}{2\epsilon_0} [-\cos(u)]_0^{\pi/4}$

   $E_x = \frac{\sigma}{2\epsilon_0} (\cos(\pi/4) - \cos(0)) = \frac{\sigma}{2\epsilon_0} \left(\frac{1}{\sqrt{2}} - 1\right)$

   $E_x = -\frac{\sigma}{2\sqrt{2}\epsilon_0} (\sqrt{2}-1)$.

5. Net Force: The net electric force on charge $q$ is $F = qE_x$.

   $F = q \left(-\frac{\sigma}{2\sqrt{2}\epsilon_0} (\sqrt{2}-1)\right)$.

   The magnitude of the force is $F = \frac{q\sigma}{2\sqrt{2}\epsilon_0} (\sqrt{2}-1)$.