The potential (in volts) of a charge distribution is given by. V(z) = 30 - 5z2 for |z| ≤ 1 m. V(z) = 35 - 10 |z| for |z| ≥ 1 m. V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ₀ (in units of ε₀) which is spread over a certain region, then choose the correct statement. (a) ρ₀ = 10 ε₀ for |z| ≤ 1 m and ρ₀ = 0 elsewhere (b) ρ₀ = 20 ε₀ in the entire region (c) ρ₀ = 40 ε₀ in the entire region (d) ρ₀ = 20 ε₀ for |z| ≤ 1 m and ρ₀ = 0 elsewhere

Answer: (a). Ez = -dV/dz. For |z| ≤ 1: Ez = 10z, so (ρ₀)/(ε₀) = (dEz)/(dz) = 10 ⇒ ρ₀ = 10ε₀. For |z| ≥ 1: |Ez| = 10 constant, so (dEz)/(dz) = 0 ⇒ ρ₀ = 0. So the answer is (a).

The potential (in volts) of a charge distribution is given by. V(z) =… | JEE Main 2016 (9 Apr, Shift 1)

Q101JEE Main 2016 · 9 APR · SHIFT 1Difficult

Electric Potential & Equipotentials

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

V(z) = 30 $-$ 5z2 for $|z| \le$ 1 m.

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

V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume $\rho_0$ (in units of $\varepsilon_0$ ) which is spread over a certain region, then choose the correct statement.

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Answer: (a)

$E_z = -\dfrac{dV}{dz}$ .
For $|z| \le 1$ : $E_z = 10z$ , so $\dfrac{\rho_0}{\varepsilon_0} = \dfrac{dE_z}{dz} = 10 \Rightarrow \rho_0 = 10\varepsilon_0$ .
For $|z| \ge 1$ : $|E_z| = 10$ constant, so $\dfrac{dE_z}{dz} = 0 \Rightarrow \rho_0 = 0$ .
So the answer is (a).

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