Electric force on the charge kept at the centre of a metallic shell

[Pushoam]

So what is the argument that says that the picture as shown below cannot be the case? Try to be as specific as you can.

Using $\oint \vec E \cdot d \vec l =0$ for the closed loop ABCA shown in the figure, where the part AB inside the cavity is along the direction of electric field, to calculate the magnitude of $\vec E$ inside the cavity, $$\oint \vec E \cdot d \vec l =0 $$$$ \int_A^B \vec E \cdot d \vec l + {\int_B^A }_{through~ C} \vec E \cdot d \vec l = 0$$$$ \int_A^B \vec E \cdot d \vec l = -{\int_B^A }_{~ through~ C} \vec E \cdot d \vec l $$

Now, ${\int_B^A }_{through~ C} \vec E \cdot d \vec l = 0$ as electric field inside the meat of conducting shell is 0.
Hence, $$ \int_A^B \vec E \cdot d \vec l = -{\int_B^A }_{through~ C} \vec E \cdot d \vec l = 0 ... (1)$$ As the part AB inside the cavity is along the direction of electric field, $$ \int_A^B \vec E \cdot d \vec l =\int_A^B E dl =0 $$$$\Rightarrow E = 0 \Rightarrow \vec E = \vec 0 $$

Hence, the above calculation and precisely, $\oint \vec E \cdot d \vec l =0$ says that there can be no electric field inside the shell.

 

[kuruman]

Yes, that is the argument. Simply put without line integrals, electric field lines point from high to low electric potential. If there were an electric field line in the cavity from point A to point B, that would mean that point A is at higher potential than B. This contradicts the fact that all points on the conductor are at the same potential.

 

[Pushoam]

Thanks for guiding me.