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PhyMedAdv 2015 • Paper 2

Electric Field Inside Off Center Spherical Cavity

A uniform spherical charge distribution has radius R1R_1 and is centered at origin OO. A spherical cavity having radius R2R_2 centered at point PP with separation OP=a=R1−R2OP = a = R_1 - R_2 is created within it (refer to the figure). If E⃗(r⃗)\vec{E}(\vec{r}) represents the electric field inside this cavity at position r⃗\vec{r}, select the correct option(s):

Question Diagram 1

Options

A

E⃗\vec{E} is uniform, its magnitude is independent of R2R_2 but its direction depends on r⃗\vec{r}

B

E⃗\vec{E} is uniform, its magnitude depends on R2R_2 and its direction depends on r⃗\vec{r}

C

E⃗\vec{E} is uniform, its magnitude is independent of aa but its direction depends on a⃗\vec{a}

D

E⃗\vec{E} is uniform and both its magnitude and direction depend on a⃗\vec{a}

Correct

Topics & Concepts

Electric Charges and FieldsElectric FieldGauss's Law

Step-by-Step Solution

To find the electric field inside the off-center spherical cavity, we can use the principle of superposition.

Let the uniform volume charge density of the solid sphere be ρ\rho.

  1. Field due to a solid sphere centered at origin OO: Consider a solid sphere of radius R1R_1 with uniform charge density ρ\rho, centered at the origin OO. Using Gauss's Law, the electric field E⃗1\vec{E}_1 at any point with position vector r⃗\vec{r} relative to OO inside the sphere is given by: E⃗1(r⃗)=ρr⃗3ε0\vec{E}_1(\vec{r}) = \frac{\rho \vec{r}}{3\varepsilon_0}

  2. Field due to the spherical cavity: Creating a cavity centered at point PP (with position vector OP⃗=a⃗\vec{OP} = \vec{a}) is equivalent to superimposing a smaller sphere of radius R2R_2 with a charge density of −ρ-\rho centered at PP.

    The position vector of the point inside the cavity relative to PP is r⃗′=r⃗−a⃗\vec{r}' = \vec{r} - \vec{a}.

    The electric field E⃗2\vec{E}_2 produced by this sphere of negative charge density at the same point is: E⃗2(r⃗)=−ρr⃗′3ε0=−ρ(r⃗−a⃗)3ε0\vec{E}_2(\vec{r}) = \frac{-\rho \vec{r}'}{3\varepsilon_0} = \frac{-\rho (\vec{r} - \vec{a})}{3\varepsilon_0}

  3. Net Electric Field inside the cavity: By the principle of superposition, the net electric field E⃗(r⃗)\vec{E}(\vec{r}) at any point inside the cavity is: E⃗(r⃗)=E⃗1(r⃗)+E⃗2(r⃗)\vec{E}(\vec{r}) = \vec{E}_1(\vec{r}) + \vec{E}_2(\vec{r}) E⃗(r⃗)=ρr⃗3ε0−ρ(r⃗−a⃗)3ε0\vec{E}(\vec{r}) = \frac{\rho \vec{r}}{3\varepsilon_0} - \frac{\rho (\vec{r} - \vec{a})}{3\varepsilon_0} E⃗(r⃗)=ρa⃗3ε0\vec{E}(\vec{r}) = \frac{\rho \vec{a}}{3\varepsilon_0}

  4. Conclusion:

    • The expression E⃗=ρa⃗3ε0\vec{E} = \frac{\rho \vec{a}}{3\varepsilon_0} is independent of the position vector r⃗\vec{r}, meaning that the electric field inside the cavity is uniform throughout its volume.
    • The magnitude of the electric field is E=ρa3ε0E = \frac{\rho a}{3\varepsilon_0}, which depends on the magnitude of the displacement vector a=∣a⃗∣a = |\vec{a}|.
    • The direction of the electric field is parallel to the vector a⃗\vec{a}.

Thus, the electric field E⃗\vec{E} is uniform and both its magnitude and direction depend on a⃗\vec{a}.

Correct Option: D

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