The net electric field at point D is determined as 3.95 x 10⁷ N/C.
Electric field at D due to charge A
E = kq/r²
where;
- r is the distance between A and D
- q is charge A
E(AD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)
E(AD) = 2.16 x 10⁷ i N/C
Electric field at D due to charge B
E = kq/r²
where;
- r is the distance between A and B
- q is charge B
r² = 5² + 5²
r² = 50
r = √50
r = 7.07 cm
E(BD) = (9 x 10⁹ x 5 x 10⁻⁶)/(0.0707²)
E(BD) = 9 x 10⁶ N/C
in x - direction = 9 x 10⁶ N/C x cos(45) = 6.36 x 10⁶ i N/C
in y - direction = 9 x 10⁶ N/C x sin(45) = 6.36 x 10⁶j N/C
Electric field at D due to charge C
E = kq/r²
where;
- r is the distance between C and D
- q is charge C
E(CD) = (9 x 10⁹ x 6 x 10⁻⁶)/(0.05²)
E(CD) = 2.16 x 10⁷ j N/C
Net electric field in x direction
Ei = 2.16 x 10⁷ i N/C + 6.36 x 10⁶ i N/C
Ei = 2.796 x 10⁷ i N/C
Net electric field in y direction
Ej = 2.16 x 10⁷ j N/C + 6.36 x 10⁶j N/C
Ej = 2.796 x 10⁷ j N/C
Resultant electric field at D
E = √Ei² + Ej²
E = √[(2.796 x 10⁷)² + (2.796 x 10⁷)²]
E = 3.95 x 10⁷ N/C
Thus, the net electric field at point D is determined as 3.95 x 10⁷ N/C.
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