The region between two concentric spheres of radii a and b (b > a) is filled with a charge such that the charge density depends only on the radial distance r, as ρ(r)=cr.. Also, there is a point charge Q placed at the center of the spheres. If the field in the region a < r < b is constant in magnitude, then c equals:

The charge in the shaded region (see figure) isq=∫ar cx4πx2dx=2πcr2−a2Field at distance r isE(r)=K(Q+q)r2=KQ+2πcr2−a2r2=KQ−2πa2c1r2+K(2πc)The above function will be independent of r ifQ−2πa2c=0⇒c=Q2πa2

The region between two concentric spheres of radii a and b (b > a) is filled with a charge such that the charge density depends only on the radial distance r, as ρ(r)=cr.. Also, there is a point charge Q placed at the center of the spheres. If the field in the region a < r < b is constant in magnitude, then c equals:

The charge in the shaded region (see figure) isq=∫ar cx4πx2dx=2πcr2−a2Field at distance r isE(r)=K(Q+q)r2=KQ+2πcr2−a2r2=KQ−2πa2c1r2+K(2πc)The above function will be independent of r ifQ−2πa2c=0⇒c=Q2πa2

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The region between two concentric spheres of radii a and b (b > a) is filled with a charge such that the charge density depends only on the radial distance r, as $ρ (r) = \frac{c}{r} .$ . Also, there is a point charge Q placed at the center of the spheres. If the field in the region a < r < b is constant in magnitude, then c equals:

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$\frac{Q}{2 π a^{2}}$

$\frac{Q}{2 π (b^{2} - a^{2})}$

$\frac{Q}{2 π b^{2}}$

$\frac{Q}{2 π a b}$

answer is A.

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Detailed Solution

The charge in the shaded region (see figure) is

$q = \int_{a}^{r} \frac{c}{x} (4 π x^{2} d x) = 2 π c (r^{2} - a^{2})$

Field at distance r is

$\begin{array}{l} E (r) = \frac{K (Q + q)}{r^{2}} = K (\frac{Q + 2 π c (r^{2} - a^{2})}{r^{2}}) \\ = K (Q - 2 π a^{2} c) \frac{1}{r^{2}} + K (2 π c) \end{array}$