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Electrostatics question

2016 · 9 Apr · Shift 1 · Q54
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Electrostatics question

2016 · 9 Apr · Shift 1 · Q54

JEE MainPhysicsElectrostaticsMCQ+4 / −1
The potential (in volts) of a charge distribution is given by. V(z) = 30 −-− 5x2 for ∣z∣\left| z \right|∣z∣≤\le≤ 1 m. V(z) = 35 −-− 10 ∣z∣\left| z \right|∣z∣ for ∣z∣\left| z \right|∣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 ρ0{\rho _0}ρ0​(in units of ε0{\varepsilon _0}ε0​) which is spread over a certain region, then choose the correct statement.
  1. A
    ρ0{\rho _0}ρ0​= 10 ε0{\varepsilon _0}ε0​ for ∣z∣\left| z \right|∣z∣ ≤\le≤ 1 m and ρ0=0{\rho _0} = 0ρ0​=0 elsewhere
  2. B
    ρ0{\rho _0}ρ0​= 20 ε0{\varepsilon _0}ε0​ in the entire region
  3. C
    ρ0{\rho _0}ρ0​= 40 ε0{\varepsilon _0}ε0​ in the entire region
  4. D
    ρ0{\rho _0}ρ0​= 20 ε0{\varepsilon _0}ε0​ for ∣z∣\left| z \right|∣z∣ ≤\le≤ 1 m and ρ0=0{\rho _0} = 0ρ0​=0 elsewhere
View written solutionFree

Correct answer: A

  1. Given potential

The potential depends only on zzz:

V(z)=30−5z2for ∣z∣≤1V(z)=30-5z^2 \quad \text{for } |z|\le 1V(z)=30−5z2for ∣z∣≤1 V(z)=35−10∣z∣for ∣z∣≥1V(z)=35-10|z| \quad \text{for } |z|\ge 1V(z)=35−10∣z∣for ∣z∣≥1

Since VVV is independent of xxx and yyy, we use the one-dimensional form of Poisson's equation:

∇2V=d2Vdz2=−ρε0\nabla^2 V=\frac{d^2V}{dz^2}=-\frac{\rho}{\varepsilon_0}∇2V=dz2d2V​=−ε0​ρ​

So,

ρ=−ε0d2Vdz2\rho=-\varepsilon_0\frac{d^2V}{dz^2}ρ=−ε0​dz2d2V​

  1. Region 1: ∣z∣≤1|z|\le 1∣z∣≤1

Here,

V(z)=30−5z2V(z)=30-5z^2V(z)=30−5z2

Differentiate:

dVdz=−10z\frac{dV}{dz}=-10zdzdV​=−10z

d2Vdz2=−10\frac{d^2V}{dz^2}=-10dz2d2V​=−10

Now apply Poisson's equation:

ρ=−ε0(−10)=10ε0\rho=-\varepsilon_0(-10)=10\varepsilon_0ρ=−ε0​(−10)=10ε0​

Thus, for ∣z∣≤1|z|\le 1∣z∣≤1,

ρ=10ε0\rho=10\varepsilon_0ρ=10ε0​

  1. Region 2: ∣z∣≥1|z|\ge 1∣z∣≥1

Here,

V(z)=35−10∣z∣V(z)=35-10|z|V(z)=35−10∣z∣

For z>1z>1z>1:

V=35−10zV=35-10zV=35−10z Rightarrow \frac{dV}{dz}=-10, \quad \frac{d^2V}{dz^2}=0

For $z<-1$: $$|z|=-z \Rightarrow V=35+10z$$ $$\frac{dV}{dz}=10, \quad \frac{d^2V}{dz^2}=0$$ Hence in both outer regions, $$\frac{d^2V}{dz^2}=0 \Rightarrow \rho=0$$ So, for $|z|\ge 1$, $$\rho=0$$ --- 4. **Check continuity at $z=\pm 1$** At $z=1$: Inside: $$V(1)=30-5=25$$ Outside: $$V(1)=35-10=25$$ At $z=-1$: Inside: $$V(-1)=30-5=25$$ Outside: $$V(-1)=35-10=25$$ Potential is continuous. Also electric field $E_z=-dV/dz$: At $z=1$: Inside slope $=-10$, outside slope $=-10$ At $z=-1$: Inside slope $=10$, outside slope $=10$ So no surface charge is present at the boundaries, consistent with only volume charge in $|z|\le 1$. --- 5. **Match with options** We found: $$\rho_0=10\varepsilon_0 \quad \text{for } |z|\le 1$$ $$\rho_0=0 \quad \text{elsewhere}$$ This matches **Option A**. --- 6. **Comparison with stored answer** Stored correct answer: **A** My derived answer: **A** They agree.
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