Question

Charge is distributed within a sphere of radius $R$ with a volume charge density $p (r) = \frac{A}{r^2} e^{-2r/a}$ where $A$ and $a$ are constants. If $Q$ is the total charge of this charge distribution, the radius $R$ is :

• A. $\frac{a}{2} \log\left(1- \frac{Q}{2\pi aA}\right) $
• B. $a \log\left(1- \frac{Q}{2 \pi aA}\right) $
• C. $a \log\left(\frac{1}{1- \frac{Q}{2\pi aA}}\right)$
• D. \(\frac{a}{2} \log\left( \frac{1}{1- \frac{Q}{2\pi aA}}\right)\)

Answer 1

The Correct Option is D

To find the radius \( R \) for which the total charge \( Q \) is distributed within the sphere, we need to integrate the given volume charge density \( p(r) = \frac{A}{r^2} e^{-2r/a} \) over the volume of the sphere.

The formula to find the total charge within a sphere with a radial charge density distribution is:

Q = \int_{0}^{R} p(r) \cdot 4\pi r^2 \, dr \

Substituting for \( p(r) \),

Q = \int_{0}^{R} \left( \frac{A}{r^2} e^{-2r/a} \right) \cdot 4\pi r^2 \, dr \

Simplifying,

Q = 4\pi A \int_{0}^{R} e^{-2r/a} \, dr \

The integral of \( e^{-2r/a} \) can be evaluated as:

\int e^{-2r/a} \, dr = -\frac{a}{2} e^{-2r/a} + C\

Applying the limits from \( 0 \) to \( R \):

Q = 4\pi A \left[ -\frac{a}{2} e^{-2r/a} \right]_{0}^{R}

This evaluates to:

Q = 4\pi A \left( -\frac{a}{2} \left( e^{-2R/a} - 1 \right) \right)

Simplifying further,

Q = 2\pi a A \left( 1 - e^{-2R/a} \right)

Solving for \( e^{-2R/a} \):

e^{-2R/a} = 1 - \frac{Q}{2\pi aA}\

Taking the natural logarithm on both sides,

-\frac{2R}{a} = \log\left(1 - \frac{Q}{2\pi aA}\right)\

Therefore,

R = -\frac{a}{2} \log\left(1 - \frac{Q}{2\pi aA}\right)\

Since the logarithmic function handles the term difference correctly, the expression simplifies further:

R = \frac{a}{2} \log\left( \frac{1}{1 - \frac{Q}{2\pi aA}} \right)\

Hence, the correct answer is \(\frac{a}{2} \log\left( \frac{1}{1- \frac{Q}{2\pi aA}}\right)\), which matches with Option 4.