What is the x-component of the total electric field due to these two charges at a point on the x-axis?
1 answer:
Find the total x - component of the electric field at that point and get the resultant or the algebraic sum.
- The electric field is a vector as it gives the force on a unit positive test charge placed at some distance from a charge.
- The electric field due to each charge at that point on the x-axis can be found using the Coulombs formula:
.
- Then one needs to get the x-component of the electric field at the point due to each charge using the concept of the resolution of forces.
- Once the x-components are found, one needs to get the resultant or the algebraic sum of them to calculate the x-component of the total electric field.
- For extra knowledge, how to find the components of a given vector can be found here: brainly.com/question/28524840
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Answer:
a) the magnitude of r is 184.62
b) the direction is 37.74° south of the negative x-axis
Explanation:
Given the data in the question;
as illustrated in the image blow;
To find the the magnitude of r, we will use the Pythagoras theorem
r² = y² + x²
r = √( y² + x²)
we substitute
r = √((-113)² + (-146)²)
r = √(12769 + 21316 )
r = √(34085 )
r = 184.62
Therefore, the magnitude of r is 184.62
To find its direction, we need to find ∅
from SOH CAH TOA
tan = opposite / adjacent
tan∅ = -113 / -146
tan∅ = 0.77397
∅ = tan⁻¹( 0.77397 )
∅ = 37.74°
Therefore, the direction is 37.74° south of the negative x-axis
