This question is incomplete, the complete question is;
A positive charge of magnitude Q1 = 6.5 nC is located at the origin.
A negative charge Q2 = -3.5 nC is located on the positive x-axis at x = 16.5 cm from the origin. The point P is located y = 10.5 cm above charge Q2.
Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.
Answer:
the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C
Explanation:
Given the data in the question;
Q1 = 6.5 nC, Q2 = -3.5 nC
from the image below, to get our angle ∅
tan∅ = opp/adj
tan∅ = 10.5 / 16.5
tan∅ = 0.636363
∅ = tan⁻¹( 0.636363 )
∅ = 32.47°
also, r1 = √( 16.5² + 10.5²)
r1 = √( 272.25 + 110.25 )
r1 = √382.5
r1 = 19.55 cm = 0.1955 m
Now, the x-component of the electric field at point P due to charge Q1 will be;
Ex = E2cos32.47°
= (kQ1/r1²)cos32.47°
we know that; k is Coulomb's law constant ( 9 × 10⁹ N.m²/ C²
Q1 = 6.5 nC = 6.5 × 10⁻⁹ C
so we substitute
= ((9 × 10⁹ × 6.5 × 10⁻⁹) / (0.1955)²) cos32.47°
= (58.5 / 0.03822025) × 0.843672
= 1291.33 N/C
Therefore, the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C
