Answered: Part A.) If a charge Q is uniformly… | bartleby

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Part A.) If a charge Q is uniformly spread throughout the volume of an insulating sphere of radius R, what fraction of the charge lies within radius R/2?

Part B.) If a charge Q is spread uniformly throughout the volume of an insulating cylinder of radius R and length L, what fraction of the charge lies with radius R/2?

Hint: the answer is not 1/2 for both part A and B.

Expert Solution

Step 1

a)

The total charge on the sphere is $Q$

and the radius of the sphere is R

According to the question, the charge is uniformly distributed throughout the volume

Then volume charge density will be

$\begin{array}{rcl} ρ & = & \frac{T o t a l c h a r g e}{v o l u m e} \\ = & \frac{Q}{\frac{4}{3} {πR}^{3}} \end{array}$

Now we have to calculate the charge present in the sphere having the radius is $r = \frac{R}{2}$

And the volume of this part is

$\begin{array}{rcl} V & = & \frac{4}{3} {πr}^{3} \\ = & \frac{4}{3} π \times {(\frac{R}{2})}^{3} \\ = & \frac{4}{3} {πR}^{3} \times (\frac{1}{8}) \end{array}$

So the charge inside this volume will be

$\begin{array}{rcl} q & = & ρ V \\ = & \frac{Q}{\frac{4}{3} {πR}^{3}} \times \frac{4}{3} {πR}^{3} \times \frac{1}{8} \\ = & \frac{1}{8} Q \end{array}$

Hence, ${\frac{1}{8}}^{t h}$ part of the total charge is present inside the volume has the radius $\frac{R}{2}$ .

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