What is the magnitude and direction of the electric field 1cm above a 24 micro columb charge? (2.16 * 10^9)N/C
Answers
Answer:
The magnitude of the electric field is simply defined as the force per charge on the test charge. ... Since electric field is defined as a force per charge, its units would be force units divided by charge units.
Answer:
The magnitude and the direction of the electric field are 2.16 × 10⁹ N/C and outwards from the point charge object.
Explanation:
Given,
Charge of the point object (q) = 24 μC
Distance from the point charge where the electric field to be found (r) = 1 cm.
To find,
The magnitude and the direction of the electric field (E)
Concept,
The electric field of a point charge (q) object at a distance 'r' is given by:
E = kq/r²
Where k = electrostatic constant = 9 × 10⁹ Nm²C⁻²
Calculation,
Using equation (1) we get:
E = (9 × 10⁹)Nm²C⁻² × (24 × 10⁻⁶)C/(1 × 10⁻²)²m²
⇒ E = 2.16 × 10⁹ N/C
As the given charge is positively charged, then the field lines will be outwards from the point charge.
Therefore, the magnitude and the direction of the electric field are 2.16 × 10⁹ N/C and outwards from the point charge object.
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