Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

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.

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

car a and b are moving in opposite directions

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d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)