compute the electric field experienced by a test a charge q = +0.08µC from a source charge q = +15µC in a vacuum when the test c
1 answer:
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Answer:
<h2>Magnitude of the second charge isExplanation:
According to columbs law;
F =
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying
Answer:
<h3>The answer is 2.51 s</h3>Explanation:
The time taken can be found by using the formula
d is the distance
v is the velocity
From the question we have
We have the final answer as
<h3>2.51 s</h3>Hope this helps you