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Three infinitely charged sheets are kept...

Three infinitely charged sheets are kept parallel to x - y plane having charge densities as shown. Then the value of electric field at 'P' is

A. `(-4sigma)/(in_(0))hatk`
B. `(4sigma)/(in_(0))hatk`
C. `(-2sigma)/(in_(0))hatk`
D. `(2sigma)/(in_(0))hatk`

A

`(-4sigma)/(in_(0))hatk`

B

`(4sigma)/(in_(0))hatk`

C

`(-2sigma)/(in_(0))hatk`

D

`(2sigma)/(in_(0))hatk`

Text Solution

AI Generated Solution

The correct Answer is:
To find the electric field at point P due to three infinitely charged sheets with given charge densities, we can follow these steps: ### Step-by-Step Solution: 1. **Identify Charge Densities**: - Let the charge densities of the three sheets be: - Sheet 1: \( \sigma \) (positive) - Sheet 2: \( -2\sigma \) (negative) - Sheet 3: \( -\sigma \) (negative) 2. **Determine Electric Field from Each Sheet**: - The electric field \( E \) due to an infinite sheet with surface charge density \( \sigma \) is given by: \[ E = \frac{\sigma}{2\epsilon_0} \] - The direction of the electric field due to a positively charged sheet is away from the sheet, while for a negatively charged sheet, it is towards the sheet. 3. **Calculate Electric Field Contributions**: - **From Sheet 1** (charge density \( \sigma \)): - \( E_1 = \frac{\sigma}{2\epsilon_0} \) (away from the sheet, in the positive z-direction) - **From Sheet 2** (charge density \( -2\sigma \)): - \( E_2 = -\frac{-2\sigma}{2\epsilon_0} = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \) (towards the sheet, in the negative z-direction) - **From Sheet 3** (charge density \( -\sigma \)): - \( E_3 = -\frac{-\sigma}{2\epsilon_0} = \frac{\sigma}{2\epsilon_0} \) (towards the sheet, in the negative z-direction) 4. **Determine the Net Electric Field**: - The net electric field \( E_{\text{net}} \) at point P is the vector sum of the electric fields from all three sheets: \[ E_{\text{net}} = E_1 + E_2 + E_3 \] - Substituting the values: \[ E_{\text{net}} = \frac{\sigma}{2\epsilon_0} - \frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} \] - Simplifying this: \[ E_{\text{net}} = \frac{\sigma}{2\epsilon_0} - \frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = \frac{\sigma - 2\sigma - \sigma}{2\epsilon_0} = \frac{-2\sigma}{2\epsilon_0} = \frac{-\sigma}{\epsilon_0} \] 5. **Direction of the Electric Field**: - Since the net electric field is negative, it indicates that the direction is in the negative z-direction (i.e., \( -\hat{k} \)). 6. **Final Result**: - Therefore, the electric field at point P is: \[ E_{\text{net}} = -\frac{2\sigma}{\epsilon_0} \hat{k} \] ### Conclusion: The correct answer is option **C**: \( -\frac{2\sigma}{\epsilon_0} \hat{k} \).
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