- #1
curious_mind
- 41
- 9
- Homework Statement
- Given a conducting metallic Hollow sphere as shown in the figure. A point charge is placed at distance ## p ## from its center. What is the magnitude of electric field at a point away at distance ## a ## from its center on the same line, as shown in the figure?
- Relevant Equations
- The electric field inside the conducting metallic hollow sphere is zero.
I have read Griffiths' Chapter 2 sections on Conductors. According to it, (if I understood it correctly) if the charge is put inside the cavity of a conductor, then the equal and opposite total charge will be induced surrounding the cavity. This charge and the total charge induced surrounding the cavity will be such that the total electric field will be canceled out inside the conductor. Then the total charge will be induced again on the surface of the conductor. So that the field outside the conductor will be due to a point charge placed inside the cavity.
But I do NOT think this argument applies to the hollow conducting sphere. Because due to point charge ##q## placed inside the hollow conducting sphere, equal and opposite total charge ##-q## will be induced to the surface of the hollow conducting sphere, is it correct? So I think the electric field inside the hollow metallic conducting sphere will NO LONGER be 0. (Is this reasoning correct, or where Is it mistaken ?)
So, if we take Gaussian surface covering a hollow sphere and a point charge with a radius greater than the given sphere's radius, then the total charge enclosed will be zero.
$$ \oint_{S} \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \dfrac{Q_{enc}}{\varepsilon_0} = \dfrac{q-q}{\varepsilon_0} = 0$$
So, I think at any point external to the given hollow sphere and point charge inside it, electric field should be ZERO.
But, The answer given as ## \vec{\mathbf{E}}_{outside} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q}{a^2} \hat{r}##. Remarkable thing is, the direction also given away from the sphere, means the field will be of repulsive nature. It is not due to induced negative charge then, otherwise direction would be opposite. I do not understand how it comes, can anyone help me? Is the answer given WRONG ?
But I do NOT think this argument applies to the hollow conducting sphere. Because due to point charge ##q## placed inside the hollow conducting sphere, equal and opposite total charge ##-q## will be induced to the surface of the hollow conducting sphere, is it correct? So I think the electric field inside the hollow metallic conducting sphere will NO LONGER be 0. (Is this reasoning correct, or where Is it mistaken ?)
So, if we take Gaussian surface covering a hollow sphere and a point charge with a radius greater than the given sphere's radius, then the total charge enclosed will be zero.
$$ \oint_{S} \vec{\mathbf{E}} \cdot d\vec{\mathbf{A}} = \dfrac{Q_{enc}}{\varepsilon_0} = \dfrac{q-q}{\varepsilon_0} = 0$$
So, I think at any point external to the given hollow sphere and point charge inside it, electric field should be ZERO.
But, The answer given as ## \vec{\mathbf{E}}_{outside} = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q}{a^2} \hat{r}##. Remarkable thing is, the direction also given away from the sphere, means the field will be of repulsive nature. It is not due to induced negative charge then, otherwise direction would be opposite. I do not understand how it comes, can anyone help me? Is the answer given WRONG ?