Question

a thin metallic spherical shell of radius R carries a charge Q on its surface a point charge Q/2 is placed at its Centre c and another charge + 2 q is placed outside a shell at a distance x from the centre find the force on the charge at the centre of the cell and at the point A ,the electric flux through the shell

Answer 1

Answer:

The electric flux through the shell is $\phi =\dfrac{Q}{2\epsilon_{0}}$ and the force is $F = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{3Q^{2}}{x^{2}}$

Explanation:

Given that,

Charge at surface = Q

Charge at point A = 2Q

Charge at center C = $\dfrac{Q}{2}$

Distance to the point A from the center = x

We know that, the electric flux will be zero at the center so, the force will be also zero at the center.

Total charge on the shell is

$Q_{t} = Q +\dfrac{Q}{2}$

$Q_{t} =\dfrac{3Q}{2}$

Now, the electric flux at point A

$E = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{\dfrac{3Q}{2}}{x^{2}}$

The force at point A

$F = 2Q\times E$

$F = 2Q\times \dfrac{1}{4\pi\epsilon_{0}}\dfrac{3q}{2x^{2}}$

$F = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{3Q^{2}}{x^{2}}$

Now, the total charge enclosed by the spherical shell

$Q_{en} = \dfrac{Q}{2}$

Using Gauss's law

The total flux is

$\phi =\dfrac{Q_{en}}{\epsilon_{0}}$

$\phi =\dfrac{Q}{2\epsilon_{0}}$

Hence, The electric flux through the shell is $\phi =\dfrac{Q}{2\epsilon_{0}}$ and the force is $F = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{3Q^{2}}{x^{2}}$

Answer 2

Answer:

hope it will help .....!!!!!