JEE Challenger
More from Electric Charges and Fields
PhyMedAdv 2013 • Paper 1

Ratio of Volume Charge Densities for Zero Electric Field

Consider two non-conducting solid spheres of radii RR and 2R2R, with uniform volume charge densities ρ1\rho_1 and ρ2\rho_2 respectively, touching each other. The net electric field at a distance 2R2R from the centre of the smaller sphere, along the line connecting the centres of the two spheres, is zero. The ratio ρ1ρ2\frac{\rho_1}{\rho_2} can be

Options

A

−4-4

B

−3225-\frac{32}{25}

Correct
C

3225\frac{32}{25}

D

44

Correct

Topics & Concepts

Electric Charges and FieldsElectric FieldGauss's Law

Step-by-Step Solution

Let the center of the smaller sphere of radius RR be at the origin C1(0,0,0)C_1(0, 0, 0) and the center of the larger sphere of radius 2R2R be at C2(3R,0,0)C_2(3R, 0, 0), since the two spheres touch each other externally. The line connecting the centers lies along the xx-axis.

The total charge of the smaller sphere is: Q1=43πR3ρ1Q_1 = \frac{4}{3}\pi R^3 \rho_1

The total charge of the larger sphere is: Q2=43π(2R)3ρ2=323πR3ρ2Q_2 = \frac{4}{3}\pi (2R)^3 \rho_2 = \frac{32}{3}\pi R^3 \rho_2

A point along the line connecting the centers at a distance 2R2R from the center of the smaller sphere can be located at either x=2Rx = 2R or x=−2Rx = -2R.

Case 1: Point at x=2Rx = 2R (between the centers of the two spheres)

  • The point x=2Rx = 2R is outside the smaller sphere (r1=2R>Rr_1 = 2R > R) and inside the larger sphere, at a distance of r2=3R−2R=R<2Rr_2 = 3R - 2R = R < 2R from the center C2C_2.

  1. The electric field due to the smaller sphere at x=2Rx = 2R is: E⃗1=14πε0Q1(2R)2i^=43πR3ρ14πε0(4R2)i^=ρ1R12ε0i^\vec{E}_1 = \frac{1}{4\pi\varepsilon_0} \frac{Q_1}{(2R)^2}\hat{i} = \frac{\frac{4}{3}\pi R^3 \rho_1}{4\pi\varepsilon_0 (4R^2)}\hat{i} = \frac{\rho_1 R}{12\varepsilon_0}\hat{i}

  2. The electric field due to the larger sphere at an internal point at a distance r2=Rr_2 = R from its center is: E⃗2=ρ2r⃗23ε0=ρ2(−Ri^)3ε0=−ρ2R3ε0i^\vec{E}_2 = \frac{\rho_2 \vec{r}_2}{3\varepsilon_0} = \frac{\rho_2 (-R\hat{i})}{3\varepsilon_0} = -\frac{\rho_2 R}{3\varepsilon_0}\hat{i}

For the net electric field to be zero: E⃗1+E⃗2=0\vec{E}_1 + \vec{E}_2 = 0 ρ1R12ε0−ρ2R3ε0=0  ⟹  ρ112=ρ23\frac{\rho_1 R}{12\varepsilon_0} - \frac{\rho_2 R}{3\varepsilon_0} = 0 \implies \frac{\rho_1}{12} = \frac{\rho_2}{3} ρ1ρ2=4\frac{\rho_1}{\rho_2} = 4

Case 2: Point at x=−2Rx = -2R (away from the larger sphere)

  • The point x=−2Rx = -2R is outside both spheres:
    • Distance from the center of the smaller sphere: r1=2Rr_1 = 2R
    • Distance from the center of the larger sphere: r2=3R−(−2R)=5Rr_2 = 3R - (-2R) = 5R

  1. The electric field due to the smaller sphere at x=−2Rx = -2R is: E⃗1=−14πε0Q1(2R)2i^=−ρ1R12ε0i^\vec{E}_1 = -\frac{1}{4\pi\varepsilon_0} \frac{Q_1}{(2R)^2}\hat{i} = -\frac{\rho_1 R}{12\varepsilon_0}\hat{i}

  2. The electric field due to the larger sphere at x=−2Rx = -2R is: E⃗2=−14πε0Q2(5R)2i^=−323πR3ρ24πε0(25R2)i^=−8ρ2R75ε0i^\vec{E}_2 = -\frac{1}{4\pi\varepsilon_0} \frac{Q_2}{(5R)^2}\hat{i} = -\frac{\frac{32}{3}\pi R^3 \rho_2}{4\pi\varepsilon_0 (25R^2)}\hat{i} = -\frac{8\rho_2 R}{75\varepsilon_0}\hat{i}

For the net electric field to be zero: E⃗1+E⃗2=0\vec{E}_1 + \vec{E}_2 = 0 −ρ1R12ε0−8ρ2R75ε0=0  ⟹  ρ112=−8ρ275-\frac{\rho_1 R}{12\varepsilon_0} - \frac{8\rho_2 R}{75\varepsilon_0} = 0 \implies \frac{\rho_1}{12} = -\frac{8\rho_2}{75} ρ1ρ2=−12×875=−3225\frac{\rho_1}{\rho_2} = -\frac{12 \times 8}{75} = -\frac{32}{25}

Thus, the possible values for the ratio ρ1ρ2\frac{\rho_1}{\rho_2} are −3225-\frac{32}{25} and 44.

Correct Options: (B), (D)

PreviousNext