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The field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/False
- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The field intensity of a charge defines the impact of the charge on a ...
Answer: a
Explanation: If a test charge +q is situated at a distance r from Q, the test charge will experience a repulsive force directed radially outward from Q. Since electric field is inversely proportional to distance, thus the statement is true.
Explanation: If a test charge +q is situated at a distance r from Q, the test charge will experience a repulsive force directed radially outward from Q. Since electric field is inversely proportional to distance, thus the statement is true.
Most Upvoted Answer
The field intensity of a charge defines the impact of the charge on a ...
The statement is True.
Explanation:
The field intensity, also known as electric field strength, is a measure of the force experienced by a test charge placed in an electric field. It is defined as the force per unit positive charge. The field intensity at a point is given by the equation:
E = F/q
Where E is the field intensity, F is the force experienced by the test charge, and q is the charge of the test charge.
The field intensity is maximum at d = 0cm because at this point the test charge is placed very close to the charge creating the field. As the distance (d) between the charges increases, the force experienced by the test charge decreases, resulting in a decrease in the field intensity.
The field intensity follows an inverse square law relationship with distance. This means that as the distance increases, the field intensity decreases at a faster rate. Mathematically, the field intensity is inversely proportional to the square of the distance:
E ∝ 1/d²
This relationship can be intuitively understood by considering the spread of the electric field lines. At a closer distance, the field lines originating from the charge are more concentrated, resulting in a higher field intensity. As the distance increases, the field lines spread out, causing a decrease in the field intensity.
Therefore, the statement is true that the field intensity of a charge is maximum at d = 0cm and minimizes as d increases.
Explanation:
The field intensity, also known as electric field strength, is a measure of the force experienced by a test charge placed in an electric field. It is defined as the force per unit positive charge. The field intensity at a point is given by the equation:
E = F/q
Where E is the field intensity, F is the force experienced by the test charge, and q is the charge of the test charge.
The field intensity is maximum at d = 0cm because at this point the test charge is placed very close to the charge creating the field. As the distance (d) between the charges increases, the force experienced by the test charge decreases, resulting in a decrease in the field intensity.
The field intensity follows an inverse square law relationship with distance. This means that as the distance increases, the field intensity decreases at a faster rate. Mathematically, the field intensity is inversely proportional to the square of the distance:
E ∝ 1/d²
This relationship can be intuitively understood by considering the spread of the electric field lines. At a closer distance, the field lines originating from the charge are more concentrated, resulting in a higher field intensity. As the distance increases, the field lines spread out, causing a decrease in the field intensity.
Therefore, the statement is true that the field intensity of a charge is maximum at d = 0cm and minimizes as d increases.
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The field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/Falsea)Trueb)FalseCorrect answer is option 'A'. Can you explain this answer?
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