Question

Point charges of +8nC,-8nC and +16nC are placed at the corners A,C and D of a
square ABCD of side 0.4 m. Find the electric field strength at B.

Answer 1

Answer:

6.75 * 10⁵ N/C

Explanation:

              E = kq/r²

Field at B due to charge at A:

         ⇒ k * q / r²

         ⇒ (9*10⁹) * (8μ) / (0.4)²

         ⇒ 4.5 * 10⁵ N/C

Field at B due to charge at C:

         ⇒ k * q / r²

         ⇒ (9*10⁹) * (8μ) / (0.4)²

         ⇒ 4.5 * 10⁵ N/C

Field at B due to charge at D:

         ⇒ k * q / r²

         ⇒ (9*10⁹) * (8μ) / (0.4√2)²

         ⇒ 2.25 * 10⁵ N/C

Resultant field due to A and C at B:    (being at 90° and equal)

  ⇒ F√2     = 4.5√2 * 10⁵

Now, this result and the field by charge at D is at 90° (not equal)

 ∴ Resultant = √(4.5√2 * 10⁵)² + (2.25 * 10⁵)²

                     = 6.75 * 10⁵ N/C