Question

. An electric charge of 8.85X10-13C is placed at the centre of the sphere of radius 1m. What is the total flux linked with the sphere? How will the flux change if another equal and opposite charge is introduced at a distance of i) 0.5m from the centre and ii) 1.5m from the centre​

Answer 1

Answer:

Electric flux linked with the sphere will be $\frac{q}{s_0}$.

Explanation:

Step 1: Electric flux, $\phi=E \cdot A$

$\begin{aligned}& \phi=\frac{1}{4 \pi \varepsilon_0} \times \frac{q}{r^2} \cdot 4 \pi r^2=\frac{4 \pi q}{4 \pi \varepsilon_0} \\& =9 \times 10^9 \times 8.85 \times 10^{-13} \times 4 \times \frac{22}{7} \\& =0.1 \mathrm{NC}^{-1} \mathrm{~m}^2\end{aligned}$

We know:

An electric charge of 8.85X10-13C is placed at the centre of the sphere of radius 1m.

To find:

1. The total flux linked with the sphere

2. charge is introduced at a distance of 0.5m from the centre and

3. charge is introduced at a distance of 1.5m from the centre​

Step 1: Consider the sphere to be a Gaussian surface.

Electric flux linked with the sphere will be $\frac{q}{s_0}$ (Gauss's Law)

$t_E=\frac{8.85 \times 10^{-13}}{8.85 \times 10^{-12}}=10^{-1} \mathrm{Nm}^2 / \mathrm{C}$.

Step 2: (i) In this case the equal and opposite charge is placed inside the sphere, so total charge inside the Gaussian surface is 0 .

$\therefore \phi_E=0$.

Step 3: (ii) In this case the equal and opposite charge is placed outside the sphere, so total charge inside the Gaussian surface remains $8.85 \times 10^{-13} \mathrm{C}$.

$\text { i.e. } t_E=\frac{8.85 \times 10^{-13}}{8.85 \times 10^{-12}}=10^{-1} \mathrm{Nm}^2 / \mathrm{C}$.

Therefore, Electric flux linked with the sphere will be $\frac{q}{s_0}$.

To learn more about similar question visit:

https://brainly.in/question/54101906

https://brainly.in/question/490634

#SPJ1