Question

Four point charges are on the corners of a rectangle, 8 m by 6 m in size. What is the magnitude of the resultant electric field at the center of the rectangle, due to all four charges?

Answer 1

Answer:

E = 1.44*10⁹* Q N/C

Explanation:

• The Electric Field due to any of the charges, at the center of the rectangle, can be expressed as follows:

       $E = \frac{K*Q}{r^{2}} (1)$

       where K = 9*10⁹ N*m²/C², Q is the charge (in Coulombs) of any

       charge, and r is the distance of the charge to the center of the

       rectangle, which it is exactly half of the diagonal of the rectangle.

• If the sides of the rectangle are 8m and 6m, the diagonal of the rectangle, just applying the Pythagorean Theorem, is as follows:

       $d = \sqrt{(8m)^{2} + (6m)^{2}} = 10 m (2)$

       so, r = 5 m.

• Replacing by the givens, the electric field due to one of the charges is simply:

      $E = \frac{K*Q}{r^{2}} = \frac{9e9N*m2/C2*Q}{(5m)^{2}} = 0.36*e9*Q N/C (3)$

• Due to the symmetry of the problem, the magnitude of the Electric Field due to any of the charges will be the same (assuming all charges are equal each other), so the total electric field, in magnitude, will be just four times the one due to any of the charges, as follows:

       $E_{tot} = 4* E = 1.44e9*Q N/C (4)$