Question Detail
Question
The distance between two positive charges 6uC and 8uC is 50 cm. Calculate the electric field intensity due to each charge at a point P in between the two charges and 10cm from the 6uC charge.
Correct Answer
Option D is the correct answer.
Detailed Explanation
P is 10 cm (0.1m) from the 6uC charge, and (50-10)=40 cm (0.4m) from the 8uC charge. E(due to 6uC) = kQ/r2 = (9x109)(6x10-6)/(0.1)2 = 54000/0.01 = 5.4x106 N/C. E(due to 8uC) = (9x109)(8x10-6)/(0.4)2 = 72000/0.16 = 4.5x105 N/C.
Hint
P is 10 cm (0.1m) from the 6uC charge, and (50-10)=40 cm (0.4m) from the 8uC charge.