JEE Advanced · Physics · 17. Electrostatics
Paragraph:
The nuclear charge \((\mathrm{Ze})\) is non-uniformly distributed within a nucleus of radius \(R\). The charge density \(\rho(r)\) [charge per unit volume] is dependent only on the radial distance \(r\) from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

Question:
For \(a=0\), the value of \(d\) (maximum value of \(\rho\) as shown in the figure) is
- A \(\frac{3 Z e}{4 \pi R^3}\)
- B \(\frac{3 Z e}{\pi R^3}\)
- C \(\frac{4 Z e}{3 \pi R^3}\)
- D \(\frac{Z e}{3 \pi R^3}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 Z e}{\pi R^3}\)
Step-by-step Solution
Detailed explanation
For \(a=0\)
\(\rho(r)=\left(-\frac{d}{R} \cdot r+d\right)\)
\(\text {Now } \int_0^R\left(4 \pi r^2\right)\left(d-\frac{d}{R} r\right) d r=\) \(\text { net charge }=Z e\)
Solving this equation, we get \(d=\frac{3 Z e}{\pi R^3}\)

\(\rho(r)=\left(-\frac{d}{R} \cdot r+d\right)\)
\(\text {Now } \int_0^R\left(4 \pi r^2\right)\left(d-\frac{d}{R} r\right) d r=\) \(\text { net charge }=Z e\)
Solving this equation, we get \(d=\frac{3 Z e}{\pi R^3}\)

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