If the electric potential is given by (\chi=cxy), calculate the electric field.

**Discussion:**

$$ E_x=-\frac{\partial\chi}{\partial x}=-cy$$

$$ E_y=\frac{\partial \chi}{\partial y}=-cx$$

Hence electric field $$ \vec{E}=-c(y\hat{i}+x\hat{j})$$

If the electric potential is given by (\chi=cxy), calculate the electric field.

**Discussion:**

$$ E_x=-\frac{\partial\chi}{\partial x}=-cy$$

$$ E_y=\frac{\partial \chi}{\partial y}=-cx$$

Hence electric field $$ \vec{E}=-c(y\hat{i}+x\hat{j})$$

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Let’s discuss a problem where we will find the charge of the sphere with the help of the three charged spheres problem. Try before reading the solution.

**The Problem: **

Consider three concentric metallic spheres \(A\), \(B\) and \(C\) of radii of \(a\), \(b\), \(c\) respectively where a<b<c . \(A\) and \(B\) are connected whereas C is grounded The potential of the middle sphere \(B\) is raised to \(V\) then what is the charge on the sphere \(C\)?

**Solution:**

Three concentric metallic spheres \(A\), \(B\) and \(C\) have radii of \(a\), \(b\), \(c\) respectively where a<b<c . \(A\) and \(B\) are connected whereas C is grounded The potential of the middle sphere \(B\) is raised to \(V\).

$$ V=\frac{Kq}{b}+\frac{KQ}{c}$$ $$

\Rightarrow \frac{k(q+Q)}{c}=0$$

$$\Rightarrow q+Q=0$$

$$\Rightarrow q=-Q

$$

$$ \frac{k(-Q)}{b}+\frac{KQ}{c}=V$$

$$\Rightarrow KQ(\frac{1}{c}-\frac{1}{b})=V$$

$$ Q=\frac{bcV}{(b-c)}4\pi\epsilon_0$$

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Let’s discuss the problem where we have to find the potential of metal sphere.

**The Problem:**

A metal sphere having a radius (r_1) charged to a potential (\phi_1) is enveloped by a thin-walled conducting spherical shell of radius (r_2). Determine the potential (\phi_2) acquired by the sphere after it has been connected for a short time to the shell by a conductor.

**Solution: **

The charge (q_1) of the sphere can be determined from the relation $$ q_1=4\pi\epsilon_0r_1$$

After the connection of the sphere to the envelope, the entire charge (q_1) will flow from the sphere to the envelope and will be distributed uniformly over its surface.

Its potential (\phi_2) (coinciding with the new value of the potential of the sphere) will be

$$ \phi_2=\frac{q_1}{4\pi\epsilon_0r_2}=\phi_1\frac{r_1}{r_2}$$