Electrostatics

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Question 213 of 311

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

V(z)=30−5z2V\left(z\right)=30-5{z}^{2} for ∣z∣≤1m\left|z\right|\leq 1m

V(z)=35−10∣z∣V\left(z\right)=35-10\left|z\right| for ∣z∣≥1m\left|z\right|\geq 1m .

V(z)V\left(z\right) does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0{\rho }_{0} (in units of ϵ0{ϵ}_{0} ) which is spread over a certain region, then choose the correct statement.

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Answer: (B) ρ0=10ϵ0{\rho }_{0}=10{ϵ}_{0} for ∣z∣≤1m\left|z\right|\leq 1m and ρ0=0{\rho }_{0}=0 else where

Electric field due to given potentials is
E1=−dVdz=10∣z∣{E}_{1}=\dfrac{-dV}{dz}=10\left|z\right| for ∣z∣≤1m∣z∣\leq 1m. Here E1{E}_{1} is proportional to zz, so charge distribution is continuous for ∣z∣≤1m\left|z\right|\leq 1m

E2=−dVdz=10{E}_{2}=-\dfrac{dV}{dz}=10 for ∣z∣≥1m\left|z\right|\geq 1m. Here, E2{E}_{2} is constant.
So, using the Poisson’s equation ∇2V=−ρϵ0{\nabla }^{2}V=-\dfrac{\rho }{{\epsilon }_{0}}, we get ρ=0\rho =0.

∴∴ The source is an infinity large non-conducting thick plate.
Hence, consider a cylindrical Gaussian surface:
Applying Gauss's law for ∣z∣≤1m∣z∣\leq 1m
E×2A=ρ×2z×Aϵ0E\times 2A=\dfrac{\rho \times 2z\times A}{{\epsilon }_{0}}
⇒(10z)(2A)=ρ(2Az)ϵ0\Rightarrow \left(10z\right)\left(2A\right)=\dfrac{\rho (2Az)}{{\epsilon }_{0}}

⇒ρ0=10ϵ0\Rightarrow {\rho }_{0}=10{\epsilon }_{0} for ∣z∣≤1m\left|z\right|\leq 1m.

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