The potential (in volts) of a charge distribution is given by

$V (z) = 30 - 5 z^{2}$ for $|z| \leq 1 m$

$V (z) = 35 - 10 |z|$ for $|z| \geq 1 m$ .

$V (z)$ does not depend on x and y. If this potential is generated by a constant charge per unit volume $ρ_{0}$ (in units of $ϵ_{0}$ ) which is spread over a certain region, then choose the correct statement.

Question

The potential (in volts) of a charge distribution is given by

V (z) = 30 - 5 z^{2}

for

|z| \leq 1 m

V (z) = 35 - 10 |z|

for

|z| \geq 1 m

.

V (z)

does not depend on x and y. If this potential is generated by a constant charge per unit volume

ρ_{0}

(in units of

ϵ_{0}

) which is spread over a certain region, then choose the correct statement.

Aicharya · Accepted Answer

Electric field due to given potentials is

$E_{1} = \frac{- d V}{d z} = 10 |z|$ for $∣ z ∣ \leq 1 m$ . Here $E_{1}$ is proportional to $z$ , so charge distribution is continuous for $|z| \leq 1 m$

$E_{2} = - \frac{d V}{d z} = 10$ for $|z| \geq 1 m$ . Here, $E_{2}$ is constant.

So, using the Poisson’s equation $\nabla^{2} V = - \frac{ρ}{ε_{0}}$ , we get $ρ = 0$ .

$∴$ The source is an infinity large non-conducting thick plate.

Hence, consider a cylindrical Gaussian surface:

Applying Gauss's law for $∣ z ∣ \leq 1 m$

$E \times 2 A = \frac{ρ \times 2 z \times A}{ε_{0}}$

$\Rightarrow (10 z) (2 A) = \frac{ρ (2 Az)}{ε_{0}}$

$\Rightarrow ρ_{0} = 10 ε_{0}$ for $|z| \leq 1 m$ .

The potential (in volts) of a charge distribution is given by V z = 30 - 5 z 2 for z ≤ 1   m V z…

Solution

The potential (in volts) of a charge distribution is given by V z = 30 - 5 z 2 for z &#8804; 1 &#160; m V z…

Solution

The potential (in volts) of a charge distribution is given by V z = 30 - 5 z 2 for z ≤ 1 m V z…