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:
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:
Show Hint
- $\vec{E}_{A}=\vec{E}_{B} ; \vec{E}_{C}=\vec{E}_{D}$
- $\vec{E}_{A}>\vec{E}_{B} ; \vec{E}_{C}=\vec{E}_{D}$
\( |E_A| = |E_B|; \; E_C > E_D \)
$\left|\vec{E}_{A}\right|=\left|\vec{E}_{B}\right| ; \vec{E}_{C}>\vec{E}_{D}$
The Correct Option is C
Solution and Explanation
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) \( |E_A| = |E_B|; \; E_C > E_D \)
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