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Question:
four charges +10nC -20nC +20nC and -10nC are placed at the corners ABCD of a square ABCD of side 0.05m calculate resultant electric field at the centre of the square
Answer:

To calculate the resultant electric field at the center of the square, you can use the principle of superposition,

which states that the electric field due to multiple point charges is the vector sum of the electric fields produced by each individual charge.

Let's label the charges and the points:

  1. Charge at A: +10nC (q1)
  2. Charge at B: -20nC (q2)
  3. Charge at C: +20nC (q3)
  4. Charge at D: -10nC (q4)

The center of the square is at point O.

First, find the electric field produced by each charge at point O using the formula for the electric field due to a point charge:

Electric field due to a point charge (E) = (k * |q|) / r2

Where:

  • E is the electric field.
  • k is Coulomb's constant ≈ 8.99 x 109 N m²/C².
  • |q| is the magnitude of the charge.
  • r is the distance from the charge to the point where the electric field is being calculated.

Now, calculate the electric field due to each charge at point O:

  1. Electric field due to q1 at O: E1 = (8.99 x 109 N m²/C² * |10 x 10-9 C|) / (0.05 m)2

  2. E1 = (8.99 x 109 * 10^-9) / (0.0025) E1 = 3.596 x 106 N/C (directed towards A)

  3. Electric field due to q2 at O: E2 = (8.99 x 109 N m²/C² * |-20 x 10-9 C|) / (0.05 m)2

  4. E2 = (8.99 x 109 * 20 x 10^-9) / (0.0025) E2 = 7.192 x 106 N/C (directed towards B)

  5. Electric field due to q3 at O: E3 = (8.99 x 109 N m²/C² * |20 x 10-9 C|) / (0.05 m)2

  6. E3 = (8.99 x 109 * 20 x 10-9) / (0.0025) E3 = 7.192 x 106 N/C (directed towards C)

  7. Electric field due to q4 at O: E4 = (8.99 x 109 N m²/C² * |10 x 10-9 C|) / (0.05 m)2

  8. E4 = (8.99 x 109 * 10 x 10^-9) / (0.0025)

  9. E4 = 3.596 x 106 N/C (directed towards D)

Now, add up these electric fields as vectors to find the resultant electric field at point O:

Resultant E = E1 + E2 + E3 + E4

To calculate the direction, note that E1, E2, E3, and E4 are all along the diagonals of the square, and they have equal magnitudes.

So, they cancel each other out, and the resultant electric field at the center O is zero N/C.

Therefore, the resultant electric field at the center of the square is zero N/C.

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