Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities and respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the spher

Question

Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities $ρ_{1}$ and $ρ_{2}$ respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio $\frac{ρ_{1}}{ρ_{2}}$ can be [2013]

Smart Achievers · Accepted Answer

(2, 4)

$Electric field E_{1} due to smaller sphere 'A' at Q$

$E_{1} = \frac{1}{4 π ε_{0}} \frac{\frac{4}{3} π R^{3} ρ_{1}}{{(2 R)}^{2}}$

$Electric field E_{2}$ $due to bigger sphere 'B' at Q$

$E_{2} = \frac{1}{4 π ε_{0}} \frac{\frac{4}{3} π {(2 R)}^{3} ρ_{2}}{{(5 R)}^{2}}$

$E_{1} + E_{2} = 0$

$\frac{1}{4 π ε_{0}} \frac{\frac{4}{3} π R^{3} ρ_{1}}{{(2 R)}^{2}} + \frac{1}{4 π ε_{0}} \frac{\frac{4}{3} π {(2 R)}^{3} ρ_{2}}{{(5 R)}^{2}} = 0$

$∴$ $\frac{ρ_{1}}{ρ_{2}} = - \frac{32}{25}$

$Electric field E_{1} due to smaller sphere 'A' at P$

$E_{1} = \frac{1}{4 π ε_{0}} \frac{ρ_{1} (\frac{4}{3} π R^{3})}{{(2 R)}^{2}}$ $= \frac{1}{4 π ε_{0}} \times \frac{ρ_{1} π R}{3} = \frac{ρ_{1} R}{12 ε_{0}}$

$Electric field E_{2} due to bigger sphere 'B' at P$

$E_{2} = \frac{1}{4 π ε_{0}} \frac{(\frac{4}{3} π {(2 R)}^{3} ρ_{2}) R}{{(2 R)}^{3}}$ $= \frac{ρ_{2} R}{3 ε_{0}}$

$E_{1} = E_{2}$ $∴$ $\frac{ρ_{1} R}{12 ε_{0}} = \frac{ρ_{2} R}{3 ε_{0}}$

$\Rightarrow \frac{ρ_{1}}{ρ_{2}} = 4$

Solution

1 $- 4$

2 $- \frac{32}{25}$

3 $\frac{32}{25}$

4 $4$

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Solution

1 -4

2 -3225

3 3225

4 4

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1 $- 4$

2 $- \frac{32}{25}$

3 $\frac{32}{25}$

4 $4$