Electric Potential With Metallic Sphere

If the electric potential of the inner metal sphere is 10 volt and that of the outer shell is 5 volt, then the potential at centre will be?

Potential on the surface of inner sphere is 10V. Now, there can not be electric field inside the metal sphere, hence for taking units positive charge from the surface of the sphere to the center of the sphere no work is required. Therefore , at the center also the potential is 10V.

What is the electric potential at the centre of sphere?

If the sphere here is the solid one….then the potential at the centre will be 1.5kq/r. This can be easily seen from the formula:POTENTIAL AT A DISTANCE r FROM the centre of a solid sphere of radius R (r is less than R ) -KQ(3R^2-r^2)/2R^3For the above case r=0.I hope this will be helpful.Thanks

What will be potential and electric field at centre of a hollow metallic sphere with charge q and radius r?

Since it's a metallic sphere, it's obviously a conductor. From Gauss' law, all the charge must reside on the outer surface of the conductor. So. the electric field at any point inside the conductor is zero. This again follows from Gauss' Theorem.Since the electric field at any point inside the conductor is zero, the potential is constant throughout the volume of the cavity. What this essentially implies is that at any point inside the shell, the potential is the same. Which implies that the potential at the center of the shell is equal to the potential at any point inside it. The potential at the center of the shell is the easiest to calculate and is given byWhere q is the charge given to the shell and r is the radius of the shell.Hope this helps.

What is the potential inside a solid metal conducting sphere kept in an external electric field E=E0 in the z direction?

Thanx for the A2AAs the sphere is made of a conducting substance (here. .metal ) the electric field inside the sphere will be 0 ..Hence the potential at any point inside the metallic sphere will be same and this will be equal to the potential of the surface.Now the potential at any point in the electric field is negative of the work done by the conservative internal force or the work done by the external force when a unit positive charge is brought from infinity (reference point ) to that particular point.As external electric field is along z direction and we can bring the unit positive charge from any arbitrary direction (may be from perpendicular to the field or parallel to the field )work done will be different in different cases. …Hence no particular potential can be defined for the spherical surface. ..hence no particular potential can be defined for any point inside the sphere

Why the electric potential at every point inside a hollow charged sphere same and equal to the electric potential on its surface?

It is so because, we have gauss theorem which says that flux coming out or entering in of a surface is equal to the net charge enclosed by that surface. In the hallow sphere case if you take any spherical surface inside hallow sphere the charge enclosed by that surface is zero, since no charge present inside hallow sphere. That means flux is zero, which means electric field (flux is electric field times area) is zero. Since electric field is spacial derivative of potential, potential is constant.E = - d(V) /dxSince E=0 Potential V must be constant inside hallow sphere.

If a hollow conducting sphere is charged, why is the potential at the center of the sphere the same as the potential on its surface?

Let's start with the expression for the electric field of a charged hollow sphere. [math]
\mathbf{E(\mathbf{r})} = \left\{
\begin{array}{lr}
\frac{Q}{4\pi \epsilon_{0} r^2 } & : r > R\\
0 & : r < R
\end{array}
\right.
[/math]Gauss' law tells us that the electric field inside of a sphere is zero as any possible within the sphere closed surface does not contain any charges, whereas outside, a charged hollow sphere behaves like a point charge.The electrical potential between two points is defined as follows:[math]
\int_{a}^{b}\mathbf{E}\cdot \mathbf{dl} = V(a) - V(b)[/math]The electrical potential is always defined with respect to some other potential. For convenience, we often express electrostatic potential with respect to potential at "infinity", which we declare zero, thus[math]
\int_{r}^{\infty}\mathbf{E}\cdot \mathbf{dl} = V(r) - V(\infty)
[/math]Assuming that [math]r > R[/math], the integral evaluates to:[math]\int_{r}^{\infty}\mathbf{E}\cdot \mathbf{dl} = \frac{Q_}{4\pi\epsilon_{0}}\int_{r}^{\infty}\frac{dr}{r^2}[/math]Conversely, assuming that [math]r \leq R [/math] we get:[math]
\int_{r}^{\infty} \mathbf{E} \cdot \mathbf{dl} = \int_{r}^{R} \mathbf{E} \cdot \mathbf{dl} + \int_{R}^{\infty} \mathbf{E} \cdot \mathbf{dl} = \int_{R}^{\infty} \mathbf{E} \cdot \mathbf{dl} = \frac{1}{4 \pi \epsilon_{0} R}
[/math]In the end, we can write:[math]
V(r) = \left\{
\begin{array}{lr}
\frac{Q}{4\pi \epsilon_{0} r } & : r > R\\
\frac{Q}{4\pi\epsilon_{0}R} & : r \leq R
\end{array}
\right.
[/math]Now, of course, this integral was evaluated somewhat hastily. We assumed that the line integral can be evaluated as a simple integral without any consideration for the path of integration. However, it has been shown that E is a conservative field, which implies that if we take a line integral from A to B, it will be independent of the path connecting the two. In fact, the whole point of introducing a potential (scalar) field would be useless were E not conservative.