Question

Three charges, qA is +6.0 μC, qB is –5.0 μC, and qc is +6.0 μC, are located at the corners of a square with each side length at 5.0 cm as shown in the diagram. Calculate the electric field at point D.

Answer 1

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.

Learn more about electric field here: brainly.com/question/14372859

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