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Grade 12th passElectrostatics

Four non conducting uniformly charged sheet are placed in X-Z plane,y=a,y=2a,y=3a,y=-a & have charge densities sigma, -2sigma, 3 sigma,-sigma respectively as shown in fig. Then find out the net electric field at origin.

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Profile image of Manisha Kumari
7 Years agoGrade 12th pass
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ApprovedApproved Tutor Answer11 Months ago

To determine the net electric field at the origin due to the four non-conducting uniformly charged sheets, we need to consider the contribution of each sheet individually and then sum them up vectorially. The electric field produced by an infinite sheet of charge is given by the formula:

Electric Field from a Charged Sheet

The electric field \( E \) due to an infinite sheet with surface charge density \( \sigma \) is:

  • For a positively charged sheet: \( E = \frac{\sigma}{2\epsilon_0} \) directed away from the sheet.
  • For a negatively charged sheet: \( E = \frac{-\sigma}{2\epsilon_0} \) directed towards the sheet.

Analyzing Each Sheet

Let's analyze the contribution of each charged sheet at the origin (0, 0, 0) located in the X-Z plane:

  • Sheet at y = a (Charge density: \( \sigma \)):

    The electric field at the origin due to this sheet is directed in the negative y-direction (away from the sheet). Thus, the field is:

    \( E_1 = -\frac{\sigma}{2\epsilon_0} \hat{j} \)

  • Sheet at y = 2a (Charge density: \( -2\sigma \)):

    This negatively charged sheet will create an electric field directed towards itself, which is also in the negative y-direction. The field is:

    \( E_2 = \frac{2\sigma}{2\epsilon_0} \hat{j} = \frac{\sigma}{\epsilon_0} \hat{j} \)

  • Sheet at y = 3a (Charge density: \( 3\sigma \)):

    This positively charged sheet will create an electric field directed away from itself, which is in the positive y-direction. The field is:

    \( E_3 = \frac{3\sigma}{2\epsilon_0} \hat{j} \)

  • Sheet at y = -a (Charge density: \( -\sigma \)):

    This negatively charged sheet will create an electric field directed towards itself, which is in the positive y-direction. The field is:

    \( E_4 = \frac{\sigma}{2\epsilon_0} \hat{j} \)

Calculating the Net Electric Field

Now, we can sum up all the electric field contributions at the origin:

Net electric field \( E_{net} \) is given by:

\( E_{net} = E_1 + E_2 + E_3 + E_4 \)

Substituting the values we calculated:

\( E_{net} = -\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{\epsilon_0} + \frac{3\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} \)

Combining these terms:

  • \( E_{net} = -\frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} + \frac{3\sigma}{2\epsilon_0} \)
  • \( E_{net} = \frac{4\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} \)
  • \( E_{net} = \frac{3\sigma}{2\epsilon_0} \hat{j} \)

Final Result

The net electric field at the origin due to the four charged sheets is:

\( E_{net} = \frac{3\sigma}{2\epsilon_0} \hat{j} \)

This indicates that the net electric field at the origin points in the positive y-direction with a magnitude of \( \frac{3\sigma}{2\epsilon_0} \).