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A non-conducting spherical shell of radi...

A non-conducting spherical shell of radius R surrounds a point charge q (q at center). The electric flux through a cap of the shell of half angle `theta` is:

A

(a)`(2piq theta)/(in_(0))`

B

(b)`(q)/(2in_(0))(1-cos theta)`

C

(c)`(q(2theta))/(4pi in_(0))`

D

(d)`(q theta)/(2pi in_(0))`

Text Solution

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The correct Answer is:
To find the electric flux through a cap of a non-conducting spherical shell surrounding a point charge \( q \) at its center, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Electric Field**: The electric field \( E \) due to a point charge \( q \) at a distance \( r \) from the charge is given by: \[ E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \] Here, \( \epsilon_0 \) is the permittivity of free space. 2. **Determine the Area of the Cap**: The area \( A \) of the spherical cap with half-angle \( \theta \) on a sphere of radius \( R \) can be calculated using the formula: \[ A = 2\pi R^2 (1 - \cos \theta) \] This formula arises from integrating the area element over the cap. 3. **Calculate the Electric Flux**: The electric flux \( \Phi_E \) through the cap is given by: \[ \Phi_E = E \cdot A \] Substituting the expressions for \( E \) and \( A \): \[ \Phi_E = \left(\frac{1}{4\pi \epsilon_0} \frac{q}{R^2}\right) \cdot \left(2\pi R^2 (1 - \cos \theta)\right) \] 4. **Simplify the Expression**: Now, we can simplify the expression: \[ \Phi_E = \frac{q}{4\pi \epsilon_0} \cdot 2\pi (1 - \cos \theta) \] \[ \Phi_E = \frac{q}{2 \epsilon_0} (1 - \cos \theta) \] 5. **Final Result**: Thus, the electric flux through the cap of the shell is: \[ \Phi_E = \frac{q}{2 \epsilon_0} (1 - \cos \theta) \]

To find the electric flux through a cap of a non-conducting spherical shell surrounding a point charge \( q \) at its center, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Electric Field**: The electric field \( E \) due to a point charge \( q \) at a distance \( r \) from the charge is given by: \[ E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} ...
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