Question

figure shows a large conducting sheet having charge density sigma on both its surfaces . the surface potential of the sheet is vo with respect to earth . a conducting sphere of radius r is placed at a separation i from the sheet as shown in figure l find the charge that will appear on the sphere after it is connected to the earth

Answer 1

4πR(σL - ε₀V₀)

Answer 2

The problem describes a large conducting sheet with charge density $\sigma$ on both its surfaces, and its surface potential is $V_0$ with respect to Earth. A conducting sphere of radius $R$ is placed at a distance $L$ from the sheet and is then connected to Earth. We need to find the charge that appears on the sphere.

Let's assume the large conducting sheet is a thin sheet located at $x=0$. The charge density on each surface is $\sigma$. Thus, the total surface charge density of the sheet is $2\sigma$. The electric field produced by a large conducting sheet with total surface charge density $\Sigma_{total}$ is $E = \frac{\Sigma_{total}}{2\epsilon_0}$. In this case, $\Sigma_{total} = 2\sigma$, so the electric field outside the sheet (for $x>0$) is $E = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$. The field is directed away from the sheet if $\sigma>0$.

The potential at a distance $x>0$ from the sheet is given by $V(x) = V(0) - \int_0^x E dx$. We are given that the surface potential of the sheet is $V_0$, so $V(0) = V_0$. $V(x) = V_0 - \int_0^x \frac{\sigma}{\epsilon_0} dx = V_0 - \frac{\sigma}{\epsilon_0} x$. This formula is valid for $x>0$.

A conducting sphere of radius $R$ is placed at a separation $L$ from the sheet. The figure indicates that $L$ is the distance from the sheet to the center of the sphere. So, the center of the sphere is located at $x=L$.

When the sphere is connected to Earth, its potential becomes 0. Since the sphere is a conductor in electrostatic equilibrium, the potential is constant everywhere inside and on the surface of the sphere, and this potential is 0. Let's consider the potential at the center of the sphere.

The potential at the center of the sphere (at $x=L$) is the sum of the potential due to the charged sheet and the potential due to the charge induced/present on the sphere. The potential at $x=L$ due to the charged sheet is $V_{sheet}(L) = V_0 - \frac{\sigma}{\epsilon_0} L$.

Let $q$ be the total charge that appears on the sphere. This charge $q$ distributes itself on the surface of the sphere. The potential at the center of the sphere due to the charge $q$ on the sphere (even in the presence of the external field from the sheet) is the same as the potential at the center of an isolated sphere with charge $q$, which is $\frac{q}{4\pi\epsilon_0 R}$. This is because the potential inside a conductor is constant, and the potential at the center due to the charge on the sphere is equal to the potential on the surface due to that charge, which is $kq/R$.

So, the total potential at the center of the sphere is the sum of the potential due to the sheet and the potential due to the charge on the sphere: $V_{total}(L) = V_{sheet}(L) + V_{sphere\_charge}(L)$. Since the sphere is connected to Earth, its potential is 0. Thus, the potential at its center is also 0: $V_{total}(L) = 0$.

Substituting the expressions for the potentials: $\left( V_0 - \frac{\sigma}{\epsilon_0} L \right) + \frac{q}{4\pi\epsilon_0 R} = 0$.

Now, we solve for the charge $q$: $\frac{q}{4\pi\epsilon_0 R} = \frac{\sigma}{\epsilon_0} L - V_0$. $q = 4\pi\epsilon_0 R \left( \frac{\sigma}{\epsilon_0} L - V_0 \right)$. $q = 4\pi\epsilon_0 R \left( \frac{\sigma L - \epsilon_0 V_0}{\epsilon_0} \right)$. $q = 4\pi R (\sigma L - \epsilon_0 V_0)$.

This is the charge that will appear on the sphere after it is connected to the Earth.