JEE Advance - Physics (2013 - Paper 2 Offline - No. 1)

Two non-conducting spheres of radii $${R_1}$$ and $${R_2}$$ and carrying uniform volume charge densities $$ + \rho $$ and $$ - \rho ,$$ respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region

JEE Advanced 2013 Paper 2 Offline Physics - Electrostatics Question 45 English
The electrostatic field is zero
The electrostatic potential is constant
The electrostatic field is constant in magnitude
The electrostatic field has same direction

Explanation

JEE Advanced 2013 Paper 2 Offline Physics - Electrostatics Question 45 English Explanation

Here we will use the concept of vectors and concept of electric field.

Let's take a point P in the overlapping region.

From $$\Delta OPQ$$,

By triangle law of vector addition,

$$\overrightarrow {{r_1}} - \overrightarrow {{r_2}} = \overrightarrow d \quad \text{... (1)}$$

Net electrostatic field at point P;

$${\overrightarrow E _{net}} = {{K{q_1}} \over {r_1^3}}\overrightarrow {{r_1}} = {{K{q_2}} \over {r_2^3}}\overrightarrow {{r_2}} $$

$$ \Rightarrow {\overrightarrow E _{net}} = {{K\left( {\rho {4 \over 3}\pi r_1^3} \right)\overrightarrow {{r_1}} } \over {r_1^3}} - {{K\left( {\rho {4 \over 3}\pi r_2^3} \right)\overrightarrow {{r_2}} } \over {r_2^3}}$$

$$ \Rightarrow {\overrightarrow E _{net}} = {4 \over 3}\rho \pi \times {1 \over {4\pi {\varepsilon _0}}}\left( {\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \right)$$

$$ \Rightarrow {\overrightarrow E _{net}} = {\rho \over {3{\varepsilon _0}}}\overrightarrow d $$

Hence, the electrostatic field is constant in magnitude and has same direction.

Therefore, options (C) and (D) are correct.

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