A 5 n force is acting on the charge 6μc at any point. what is magnitude of the electric field at that point?
1 answer:
The magnitude of the electric field at the given point is 8.33 * 10⁵N/C.
Given the data in the question;
- Electrostatic force exterted;
- Charge;
- Magnitude of electric field;
The magnitude of electric field is simply referred to as the force per charge on a test charge.
It is expressed as;
Here, F is electrostatic force exerted and q is the charge.
To determine the magnitude of the electric field at the given point, we substitute our given values into the expression above.
Therefore, the magnitude of the electric field at the given point is 8.33 * 10⁵N/C.
Learn more about coulomb's law: brainly.com/question/506926
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Using the binomial distribution, it is found that since 16 is more than 2.5 standard deviations above the mean, it is a unusually high number.
<h3>What is the binomial probability distribution?</h3>It is the probability of <u>exactly x successes on n repeated trials, with p probability</u> of a success on each trial.
The expected value of the binomial distribution is:
E(X) = np
The standard deviation of the binomial distribution is:
A measure is considered to be unusually high if it is more than 2.5 standard deviations above the mean.
In this problem, we hav ehtat:
- 34% of companies reject candidates because of information found on their social media, hence p = 0.34.
- 27 human resource professionals are randomly selected, hence n = 27.
Then, we find the threshold for unusually high values as follows:
E(X) = np = 27 x 0.34 = 9.18
T = 9.18 + 2 x 2.46 = 14.1.
Since 16 is more than 2.5 standard deviations above the mean, it is a unusually high number.
More can be learned about the binomial distribution at brainly.com/question/24863377
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