Question

Two infinite identical charged sheets and a charged spherical body of charge density ' $\rho$ ' are arranged as shown in figure. Then the correct relation between the electrical fields at $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D points is:

• A. $\vec{E}_{A}=\vec{E}_{B} ; \vec{E}_{C}=\vec{E}_{D}$
• B. $\vec{E}_{A}>\vec{E}_{B} ; \vec{E}_{C}=\vec{E}_{D}$
• C. \( |E_A| = |E_B|; \; E_C > E_D \)
• D. $\left|\vec{E}_{A}\right|=\left|\vec{E}_{B}\right| ; \vec{E}_{C}>\vec{E}_{D}$

Hint:

The electric field depends on the arrangement and charge density of the sheets and spherical body.

Answer 1

The Correct Option is C

We are given two infinite identical charged sheets and a charged spherical body of charge density \( \rho \). We are to find the correct relation between the electric fields at points A, B, C, and D as shown in the figure.

Concept Used:

For an infinite plane sheet of charge density \( \sigma \), the electric field near it is given by:

\[ E = \frac{\sigma}{2\varepsilon_0} \]

The direction of the field is away from the sheet if \( \sigma > 0 \) (positive charge) and toward the sheet if \( \sigma < 0 \) (negative charge).

For a uniformly charged non-conducting solid sphere of charge density \( \rho \):

• Inside the sphere (\( r < R \)):
• Outside the sphere (\( r > R \)):

Step-by-Step Solution:

Step 1: Analyze the field due to the two infinite sheets.

• Between the two sheets (region containing A and B):
• Outside the two sheets (points C and D):

Step 2: Now include the effect of the charged spherical body.

The sphere creates an additional electric field directed radially outward since it has positive charge density \( \rho \).

• At A and B (inside or near the sphere): \( E_A > E_B \) because the field increases linearly with distance from the center inside the sphere.
• At C and D (outside the sheets): Both experience the same sheet field, but the sphere’s field at D (to the right) adds to the field from the sheets, while at C (to the left), it subtracts from it.

Step 3: Combine all effects.

Hence, inside the region between sheets, both A and B have small but similar fields (mostly due to the sphere), while outside, D has a stronger net field than C.

Final Computation & Result:

Therefore, the correct relation between the electric fields is:

\[ |E_A| = |E_B|; \quad E_C > E_D \]

Final Answer: Option (3) \( |E_A| = |E_B|; \; E_C > E_D \)